# Constructing Homomorphisms via Structure-Preserving Extensions

In this post, we study methods of constructing a group homomorphism by first constructing a funciton on a generating set and extending it in a homomorphic way.

This post consists of two parts: the regular text, and the red text. You are strongly advised to skip the part in red on the first read.

## 1. Linear Extensions

We first consider a linear-algebraic analogue. Let $T:V \to W$ be a linear transformation between two vector spaces $V$ and $W$ over a field $k$. Take a basis $\{v_\alpha : \alpha \in I\}$ of $V$. Since every vector in $V$ can be written in the form $\sum_\alpha a_\alpha v_\alpha$, we see that the image of a generic vector in $V$ under $T$ is of the form

$T\left( \sum_\alpha a_\alpha v_\alpha \right) = \sum_\alpha a_\alpha T(v_\alpha)$.

In other words, the values of the linear transformation $T$ are completely determined by the values of $T$ on the basis vectors $v_\alpha$. This suggests the following:

Proposition 1.1. Let $V$ and $W$ be vector spaces over a field $k$. Assume that $\{v_\alpha : \alpha \in I\}$ is a basis of $V$. Given a function $\tau:\{v_\alpha : \alpha \in I\} \to W$, the linear extension $T:V \to W$ of $\tau$ defined by setting

is a linear transformation. Moreover, $\operatorname{im} T = \operatorname{span} (\operatorname{im} \tau)$. (See also the first universal property of free vector spaces.)

Proof. We forced $T$ to be linear, didn’t we? $\square$

Example 1.2. Let $e_x = (1,0,0)$, $e_y = (0,1,0)$, and $e_z = (0,0,1)$. The collection $\{e_x,e_y,e_z\}$ forms a basis of $\mathbb{R}^3$. We define a function $\tau:\{e_x,e_y,e_z\} \to \mathbb{R}^2$ by setting $\tau(e_x) = (1,0)$, $\tau(e_y) = (0,1)$, and $\tau(e_z) = (1,1)$. The linear extension $T:\mathbb{R}^3 \to \mathbb{R}^2$ of $\tau$ is given by the formula

so that $T|_{\{e_x,e_y,e_z\}} = \tau$. $T$ is easily seen to be linear. $\square$

Suppose we want to determine injectivity or surjectivity of a linear extension $T:V \to W$ of $\tau:\{v_\alpha : \alpha \in I\} \to W$. For this, we need additional information about how $T$ interacts with the codomain $W$. For example, if $T(\{v_\alpha: \alpha \in I\})$ spans $W$, then the linearity of $T$ implies that $T$ is surjective. Similarly, if $T(\{v_\alpha :\alpha \in I\})$ is a linearly independent set in $W$, then $T$ must be injective. We can codify this observation in the following generators-to-generators principle:

Proposition 1.3. Let $V$ and $W$ be vector spaces over a field $k$. Take bases $\{v_\alpha : \alpha \in I\}$ and $\{w_\beta : \beta \in J\}$ of $V$ and $W$, respectively. Given a function $\tau:\{v_\alpha : \alpha \in I\} \to \{w_\beta : \beta \in J\}$, the linear extension $T:V \to W$ defined in Proposition 1.1 is injective if $\tau$ is injective and is surjective if $\tau$ is surjective. (See also the second universal property of free vector spaces.)

Proof. If $\tau$ is injective, then $\operatorname{im}\tau = T(\{v_\alpha : \alpha \in I\})$ is a linearly independent subset of $W$. Since $\operatorname{im} T = \operatorname{span} (\operatorname{im} \tau)$, we see that every element of $\operatorname{im} T$ can be written uniquely as a finite linear combination of elements in $\operatorname{im} \tau$. Therefore, if $Tv = Tv’$ for some $v,v’ \in V$, then we can find $v_{\alpha_1},\ldots,v_{\alpha_n} \in \{v_\alpha : \alpha \in I\}$ and scalars $a_1,\ldots,a_n$ such that

$\displaystyle Tv = \sum_{i=1}^n a_i \tau(v_{\alpha_i}) = Tv’$.

It follows from the definition of $T$ that $v = \sum_{i=1}^n a_iv_{\alpha_i} = v’$. We conclude that $T$ is injective.

If $\tau$ is surjective, then $\operatorname{im}\tau = \{w_\beta : \beta \in J\}$ is a spanning set of $W$. Since $\operatorname{im} T = \operatorname{span} (\operatorname{im} \tau)$, we conclude that $T$ is surjective. $\square$

Example 1.4. The linear map in Example 1.2 is surjective. (Check this!) $\square$

## 2. Generators

In order to develop a group-theoretic analogue of the method of linear extensions, we must formulate a group-theoretic analogue of vector bases.

Definition 2.1. Let $G$ be a group and $S$ be a subset of $G$. The subgroup of $G$ generated by $S$ is the smallest subgroup of $G$ that contains $S$, i.e.,

$\displaystyle \langle S \rangle = \bigcap_{\substack{H \leq G \\ H \supseteq S} H}$ H,

the intersect of all subgroups of $G$ that contain $S$. If $G = \langle S \rangle$, then we say that $S$ is a generating set of $G$; in this case, elements of $S$ are said to be generators of $G$. We say that $G$ is finitely generated if there exists a finite set $S = \{s_1,\ldots,s_n\}$ such that $G = \langle S \rangle$. In this case, we often write $G = \langle s_1,\ldots,s_n \rangle$.

In order to justify the use of the term subgroup, we establish the following:

Proposition 2.2. Let $G$ be a group and $\{H_\alpha : \alpha \in I\}$ be a collection of subgroups of $G$. The intersection $\bigcap_{\alpha \in I} H_\alpha$ is a subgroup of $G$.

Proof. Since $1_G \in H_\alpha$ for each $\alpha \in I$, we see that $1_G \in \bigcap_{\alpha \in I} H_\alpha$.

To show that $\bigcap_{\alpha \in I} H_\alpha$ is closed under group operation, we fix $x,y \in \bigcap_{\alpha \in I} H_\alpha$. This implies that $x,y \in H_\alpha$ for all $\alpha \in I$. Since each $H_\alpha$ is a subgroup of $G$, we conclude that $xy \in H_\alpha$ for all $\alpha \in I$. It now follows that $xy \in \bigcap_{\alpha \in I} H_\alpha$, as was to be shown.

It remains to show that $\bigcap_{\alpha \in I} H_\alpha$ is closed under the inverse operation. To this end, we fix $x \in \bigcap_{\alpha \in I} H_\alpha$. This implies that $x \in H_\alpha$ for all $\alpha \in I$. Since each $H_\alpha$ is a subgroup of $G$, we conclude that $x^{-1} \in H_\alpha$ for all $\alpha \in I$. It now follows that $x^{-1} \in \bigcap_{\alpha \in I} H_\alpha$, as desired. $\square$

We now show that, just like vector bases, we can construct from generators any element of a group by a finitary process.

Proposition 2.3. Let $G$ be a group and $S$ be a subset of $G$. We let $S^{-1} = \{s^{-1} : s \in S\}$ and define, for each $n \in \mathbb{N}$, the set

$S^n = \{s_1 \cdots s_n : s_1,\ldots,s_n \in S \cup S^{-1}\}$.

We then have the identity

$\displaystyle \langle S \rangle = \bigcup_{n=1}^\infty S^n$.

Remark. $S^n$ signifies the collection of all $n$-fold products of elements in $S$ or their inverses. The claim, then, is that $\langle S \rangle$ corresponds to the collection of all finite products of generators and their inverses.

Proof. For notational convenience, we set $[S] = \bigcup_{n=1}^\infty S_n$. Since $\langle S \rangle$ is a group that contains $S$, it also contains $S^{-1}$. By applying the closure property repeatedly, we see that $S^n \subseteq \langle S \rangle$ for all $n \in \mathbb{N}$. It follows that $[S] \subseteq \langle S \rangle$.

To establish the reverse inclusion, we show that $[S]$ is a subgroup of $G$. $[S]$ is then a subgroup of $G$ containing $S$, whence we must have the inclusion relation $\langle S \rangle \subseteq [S]$.

It therefore suffices to show that $[S] \leq G$. To this end, we first observe that $1_G = ss^{-1} \in S_2 \subseteq [S]$ for any fixed $s \in S$. Secondly, if we are given $s,t \in [S]$, then we can find $s_1,\ldots,s_n ,t_1,\ldots,t_m \in S \cup S^{-1}$ such that $s = s_1 \cdots s_n$ and $t = t_1 \cdots t_m$. It then follows that

$st = s_1 \cdots s_n t_1 \cdots t_m \in S^{n+m} \subseteq [S]$.

Finally, if $s \in [S]$, then we can find $s_1,\ldots,s_n \in S \cup S^{-1}$ such that $s = s_1 \cdots s_n$, whence

$s^{-1} = (s_1 \cdots s_n)^{-1} = s_n^{-1} \cdots s_1^{-1} \in S^n \subseteq [S]$.

We conclude that $[S] \leq G$. $\square$

Example 2.4. We say that a group $G$ is cyclic if there exists an element $x \in G$ such that $G = \langle x \rangle$. The main example is the group of integers $\mathbb{Z}$, which is generated by $1$. For each $n \in \mathbb{N}$, the set $n \mathbb{Z} = \{nk : k \in \mathbb{Z}\}$ is a subgroup of $\mathbb{Z}$. Since $\mathbb{Z}$ is an abelian group, every subgroup of $\mathbb{Z}$ is normal, whence we can form the quotient group

$\mathbb{Z}/n\mathbb{Z} = \{[0],[1],\ldots,[n-1]\}$,

where $[m] = \{nk + m : k \in \mathbb{Z}\}$. Since $\mathbb{Z}/n\mathbb{Z} = \langle [1] \rangle$, the quotient group $\mathbb{Z}/n\mathbb{Z}$ is cyclic as well.We note that not every generating set of a cyclic group is of cardinality one. Indeed, $\mathbb{Z} = \langle 1,2\rangle = \langle 2,3 \rangle = \langle 2,5 \rangle$ and so on. $\square$

Example 2.5. Consider $S_3$, the symmetric group on three letters. We claim that $\{(12),(13),(23)\}$ is a generating set for $S_3$: see the Wikipedia page on cycle notation. To establish the claim, we first observe that $S_3 = \{1,(12),(13),(23),(123),(132)\}$. Now:

• $1 = (12)(12)$;
• $(132) = (23)(12)$;
• $(123) = (23)(13)$;

this verifies the claim.

It is a general fact that $S_n$, the symmetric group on $n$ letters, is generated by 2-cycles. $\square$

Example 2.6. (Presentation of groups via generators and relations). We define the free group on $n$ letters $s_1,\ldots,s_n$, denoted by $F_n$, as follows. We first define the symbol $1$ to be the identity element of $F_n$. For each $k$, we define the symbol $s_k^{-1}$ to be the inverse of $s_k$, and declare that $s_k s_k^{-1} = s_k^{-1} s_k = 1$. Let $S = \{s_1,\ldots,s_n\}$ and $S^{-1} = \{s_1^{-1},\ldots,s_n^{-1}\}$. For each $m \in \mathbb{N}$, we define a word of length m to be the string

where $w_1,\ldots,w_m \in S \cup S^{-1}$ such that $w_k w_{k+1} \neq 1$ for all $1 \leq k \leq m-1$. In other words, a word of length $m$ is a string of letters $s_1,\ldots,s_n$ and their inverses, written down in a way that no two adjacent symbols cancel each other out. For example, $s_1s_3s_7^{-1}s_5s_4s_3^{-1}$ is a word of length 6, whereas $s_1s_3s_3^{-1}s_7$ is actually a word of length 2, because $s_3$ and $s_3^{-1}$ cancel out.

We now define $F_n$ to be the collection of all words of finite length, and define the group product of two words $w_1 \cdots w_m$ and $w_{m+1} \cdots w_n$ to be the concatenation

of the words. In order for the concatenation to be a word, we must prune out the symbols that cancel each other out: if $w_kw_{k+1} = 1$, then we take out both $w_kw_{k+1}$. The reduced concatenation is the resulting string; if no symbol is left, then the we declare the reduced concatenation to be $1$. For example, the product of $s_1s_3s_7^{-1}$ and $s_7^{-1}s_3^{-1}s_2s_4$ is

$\color{red}{s_1s_3s_7^{-1}s_7^{-1}s_3^{-1}s_2s_4 = s_1s_3s_3^{-1}s_2s_4 = s_1s_2s_3}$,

a word of length 3. The product of $s_1^{-1}s_3$ and $s_3^{-1}s_1$ is

$\color{red}{s_1^{-1}s_3s_3^{-1}s_1 = s_1^{-1}s_1 = 1}$,

a word of length 1. We observe that the inverse of a word $w_1 \cdots w_n$ is $w_n^{-1} \cdots w_1^{-1}$. Indeed,

and

$\color{red}{(w_n^{-1} \cdots w_1^{-1})(w_1 \cdots w_n) = w_n^{-1} \cdots w_2^{-1} w_2 \cdots w_n = \cdots = w_n^{-1} w_n = 1}$.

A relation on $F_n$ is an equivalence relation on $F_n$. For example, $xy \sim yx$ for all $x,y \in F_n$ is the abelian relation on $F_n$, since we have identified $xy$ with $yx$. We define the “relation set” $N’ = \{xy(yx)^{-1} : x,y \in F_n\}$, so that $w \sim w’$ if and only if $w(w’)^{-1} \in N’$. We now let $N$ be the normal closure of $N’$, viz.,

By construction, $N$ is a normal subgroup of $F_n$, so we can construct the quotient group $F_n/N$. Since $N$ contains $N’$, the quotient group $F_n/N$ must be abelian. Indeed, given $x,y \in F_n$, the commutator $xyx^{-1}y^{-1}$ is in $N$, whence the coset $xyx^{-1}y^{-1}N$ is equal to the identity coset $N$.

In a similar manner, we can define a group generated by $S$ subject to relations $R$ as follows. Given a finite set $S$ of generators, we construct the free group $F_n$, where $n$ is the number of generators. We let $R= \{\sim_1,\ldots,\sim_m\}$ be a collection of equivalence relations on $F_n$ and define the “relation set” $N’$ by declaring $xy^{-1} \in N’$ if and only if $x \sim_k y$ for some $1 \leq k \leq m$. We then take $N$ to be the normal closure of $N’$ and define

$\color{red}{\langle S \mid R \rangle = F_n/N}$.

$\langle S \mid R \rangle$ is then the group of all words generated by $S$, with the condition that two words $w$ and $w’$ are identified in case $w \sim_k w’$ for some $k$.

It is typical to write out the relations using the $=$ symbol, rather than the $\sim$ symbol. For example, the abelian group example above can be written as $\langle s_1,\ldots,s_n \mid s_i s_j = s_j s_i \mbox{ for all } 1 \leq i,j \leq n \rangle$. This notation also suggests the following general fact. If $G$ is a group such that $G = \langle x_1,\ldots,x_n \rangle$, then we can work out the relations that the generators satisfy as follows: whenever the two products of generators are equal in $G$, the equal-relation induces an equivalence relation on $G$. We collect all such equivalent relations in $R$ and generate the set $N’$ as above. Taking the normal closure $N$ of $N’$, we see that the quotient group $F_n/N$ is essentially the same as $G$, having been generated by $n$ generators subject to the same relation. See Example 3.3 for a concrete example

We conclude with a general remark. Echoing the universal property of free vector spaces, we can also work out a universal-property characterization of free groups. Given a set $S$, we define the free group on $S$ to be the group $F_S$ containing $S$ as a subset such that each function $\phi:S \to G$ into another group $G$ admits a group homomorphism $\varphi:F_S \to G$ such that the following diagram commutes:

Here $\iota_S:S \to F_S$ is the canonical injection map $\iota_S(s) = s$. A group is said to be free if it is isomorphic to the free group generated by some set. $\mathbb{Z}$ is isomorphic to the free group generated by one letter. The symmetric group $S_n$, as well as the groups presented in the next two examples, are not free. This is in stark contrast with vector spaces, as every vector space is free. The key difference is that vector spaces admit bases, whereas groups, in general, do not. A good w way to think about freeness as an algebraic property is “it has a basis”, although this isn’t entirely accurate. $\square$

Example 2.7. $D_n$, the dihedral group of degree $n$, is defined to be $\langle \rho,\tau \mid \rho^n = 1, \tau^2 = 1, \rho\tau = \tau\rho^{n-1}\rangle$, the group generated by $\rho$ and $\tau$, subject to the relations

1. $\rho^n = 1$;
2. $\tau^2 = 1$;
3. $\rho\tau = \tau\rho^{n-1}$.

We show that $D_n$ is a group of order $2n$. We first note that every word (see Example 2.6) in $D_n$ can be written in the form $\tau^j\rho^i$; this is a consequence of relation 3. Indeed, we see, for example, that

Now, by relations 1 and 2, we have the restriction $0 \leq i \leq 1$ and $0 \leq j \leq n-1$. Therefore, there are $2 \cdot n = 2n$ unique elements of $D_n$. $\square$

Example 2.8. $Q$, the group of unit quaternions, is the group generated by $-1$, $i$, $j$, and $k$ subject to the relations

1. $(-1)^2 = 1$;
2. $i^2 = -1$;
3. $j^2 = -1$;
4. $k^2 = -1$;
5. $ij=k$;
6. $ji=-k$;
7. $jk = i$;
8. $kj = -i$;
9. $ki = j$;
10. $ik = -j$.

Here we have denoted the inverses of $i$, $j$, and $k$ by $-i$, $-j$, and $-k$, respectively.

The relations show that every word (see Example 2.6) in $Q$ is of length 1, as any multi-letter word can be reduced to a one-letter word. There are 8 unique words of length 1 in $Q$: $1$,$-1$, $i$, $-i$, $j$, $-j$, $k$, $-k$.

## 3. Homomorphic Extensions

Let us now return to the task of developing a group-theoretic analogue of the method of linear extensions. One key difference between generators of a group and basis elements of a vector spaces is that the former does not provide a unique presentation of a generic element.

Example 3.1. Consider $D_4\ = \langle \rho,\tau \mid \rho^4 = \tau^2 = 1, \rho\tau =\tau \rho^3 \rangle$, the dihedral group of degree 4 (see Example 2.7). It is true, for example, that

It is crucial to note that group elements are subject to various relations that differ from group to group (see Example 2.6), whereas all vectors are subject to the same relations. (Formally, not every group is free, but every vector space is free; see Example 2.6.) This causes naïve adaptations of the method of linear extensions to fail, as illustrated in the example below.

Example 3.2. Consider the group of integers $\mathbb{Z}$ and the group $\mathbb{Z}/3\mathbb{Z} = \{[0],[1],[2]\}$ of integers mod 3. Observe that $\mathbb{Z} = \langle 1 \rangle$ and $\mathbb{Z}/3\mathbb{Z} = \langle [1]\rangle$. The map $\varphi:\mathbb{Z}/3\mathbb{Z} \to \mathbb{Z}$ defined by setting $\varphi([1]) = 1$ and declaring

• $\varphi([2]) = \varphi([1] + [1]) = \varphi([1]) + \varphi([1]) = 2$
• $\varphi([0]) = \varphi([1]+[1]+[1]) = \varphi([1])+\varphi([1]) + \varphi([1]) = 3$

is not a group homomorphism, despite the obvious attempt to force $\varphi$ to be structure-preserving Indeed, $[1]+[1]+[1]+[1]+[1] = [2]$, whereas

Moreover, $\varphi$ does not map $[0]$, the identity element in $\mathbb{Z}/3\mathbb{Z}$, to $0$, the identity element in $\mathbb{Z}$. $\square$

In using the generators-to-generators trick, then, we must take the relations imposed on the domain group and the codomain group into consideration.

Example 3.3. The direct product $S_3 \times \mathbb{Z}/2\mathbb{Z}$ of the symmetric group on three letters $S_3$ and the cyclic group $\mathbb{Z}/2\mathbb{Z}$ is a nonabelian group, as the subgroup $S_3 \times \{[0]\}$ is nonabelian. Since $|S_3| = 6$ and $|\mathbb{Z}/2\mathbb{Z}| = 2$, the argument below shows that $|S_3 \times \mathbb{Z}/2\mathbb{Z}| = 12$.

The order of the direct product $G \times H$ of two finite groups $G$ and $H$ is $|G||H|$. Indeed, if $(g,h) \in G \times H$ and $(g_0,1_H) \in G \times \{1_H\}$, then

It follows that $G \times \{1_H\}$ is a normal subgroup of $G \times H$. Now, the mapping $\varphi: H \to (G \times H)/(G \times \{1_H\})$ defined by the formula $\varphi(h) = h(G \times \{1_H\})$ is easily seen to be a bijection. Since $|(G \times H)/(G \times \{1_H\}) = [G \times H: G \times \{1_H\}]$ and $|G \times \{1_H\}| = |G|$, we conclude from Lagrange’s theorem that

$|G \times H| = |G \times \{1_H\}|[G \times H : G \times \{1_H\}] = |G||H|,$ as was to be shown.

Note that $a = ((123),[1])$ is an element of order 6—see the Wikipedia page on cycle notation for the definition of $(123)$—as we can explicitly compute the 6 powers:

1. $( (123) , [1] )$;
2. $( (132) , [0] )$;
3. $( 1_{S_3} , [1] )$;
4. $( (123) , [0] )$;
5. $( (132) , [1] )$;
6. $( 1_{S_3}, [0] )$.

We now set $b = ( (12),[1])$, which is easily seen to be an element of order 2. Since $(12)(132) = (123)$, we see that

We claim that $\{a,b\}$ generate $S_3 \times \mathbb{Z}/2\mathbb{Z}$. To prove this, it suffices to show that $S = \{ ( (12), [0] ) , ( (13), [0] ), ( (23), [0] ), (1_{S_3}, [1])\}$ is contained in $\langle a,b \rangle$, the subgroup generated by $a$ and $b$. Assume the inclusion relation for the sake of the argument. We have seen in Example 2.5 that the 2-cycles generate $S_3$; this, in particular, implies that $S_3 \times \{[0]\}$ is contained in $\langle a,b \rangle$. Since $(1_{S_3},[1])$ is in $\langle a,b \rangle$, elements of the form $\sigma,[1]$ for arbitrary $\sigma \in S_3$ are also in $\langle a,b \rangle$. It follows that the full group $S_3 \times \mathbb{Z}/2\mathbb{Z}$ is contained in $\langle a,b \rangle$, whence $S_3 \times \mathbb{Z}/2\mathbb{Z} = \langle a,b \rangle$.

Now, $(123)(12)(132) = (23)$ and $(132)(12)(123) = (13)$, and so

• $ba^3 = ( (12),[0] )$;
• $a^2 (ba^3) a^4 ( (13) , [0] )$;
• $a^4 (ba^3) a^2 = ( (23) , [0] )$;
• $a^3 = (1_{S_3}, [1])$.

It follows that $S_3 \times \mathbb{Z}/2 \mathbb{Z}$ is generated by $a$ and $b$, which, in turn, are subject to the relations

1. $a^6 = 1$;
2. $b^2 = 1$;
3. $ab = ba^5$.

Recall now that $D_6 = \langle \rho,\tau \mid \rho^6 = \tau^2 = 1, \rho\tau = \tau \rho^5\rangle$, the dihedral group of degree 6 (see Example 2.7). Indeed, $D_6$ is a nonabelian group of order 12 generated by two elements subject to the same relations!

We define a mapping $\varphi: S_3 \times \mathbb{Z}/2\mathbb{Z} \to D_6$ by setting $\varphi(a^nb^m) = \rho^n \tau^m$ for all $n,m \in \mathbb{N}$. This is a homomorphic extension of the generators-to-generators mapping $\phi:\{a,b\} \to D_6$ given by setting $\phi(a) = \rho$, $\phi(b) = \tau$; indeed, an equivalent way of defining $\varphi$ is to declare that

for all $n,m \in \mathbb{Z}$, which is resminiscent of the linear extension definition in Proposition 1.1. To check that $\varphi$ is a group homomorphism, we fix $n_1,n_2,m_1,m_2 \in \mathbb{Z}$. Since $ba^5 = ba^{-1}$, we apply relation 3 repeatedly to conclude that

Since the exponents were arbitrary, we conclude that $\varphi$ is a group homomorphism. By restricting the exponents to $0 \leq n \leq 5$ and $0 \leq m \leq 1$, we see that $\varphi$ is a bijection. It follows that $S_3 \times \mathbb{Z} / 2\mathbb{Z} \cong D_6$. $\square$

What the above example shows is that the generators-and-generators trick for constructing homomorphisms works if two groups in question have the same number of generators that are subject to the same relations. This appears to be too stringent of a condition, however, as we expect such groups to be isomorphic, anyway. (If a group $G$ is generated by $\{x_1,\ldots,x_n\}$, subject to a set of relations $R$, then $G$ is isomorphic to $F_n/N$, where $F_n$ is the free group on $x_1,\ldots,x_n$, and $N$ is the normal closure of the set $N’$ generated by the relations $R$; see Example 2.6.)

In fact, having merely compatible relations suffices, as the next example shows.

Example 3.4. Consider $D_4 = \langle \rho, \tau \mid \rho^4 = \tau^2 = 1, \rho\tau = \tau \rho^3 \rangle$, the dihedral group of degree 4 (see Example 2.7). We note that $\langle \rho^2 \rangle$ is a normal subgroup of $D_4$. Indeed,

for all $n,m \in \mathbb{Z}$. The quotient group $D_4 / \langle \rho^2 \rangle$ is then generated by $[\rho] = \rho\langle \rho^2 \rangle$ and $[\tau] = \tau\langle \rho^2 \rangle$, which are now subject to the relations

1. $[\rho]^2 = 1$;
2. $[\tau]^2 = 1$;
3. $[\rho][\tau] = [\tau][\rho]$.

This stems from the fact that $[\rho^2] = [1]$.

Now, we know an abelian group with two generators, both of order 2: $\mathbb{Z}/2\mathbb{Z}\times \mathbb{Z}/2\mathbb{Z}$. Analogously as in Example 3.3, we can define a group homomorphism $\bar{\varphi}:D_4 / \langle \rho^2 \rangle \to \mathbb{Z}/2 \mathbb{Z} \times \mathbb{Z}/ 2 \mathbb{Z}$ by setting $\bar{\varphi}( [\rho]^n[\tau]^m ) = (n,m)$. $\bar{\varphi}$ is an isomorphism.

This suggests, however, that $\varphi:D_4 \to \mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z}$ defined by setting $\varphi(\rho^n \tau^m) = (n,m)$ should also be a group homomorphism, and it indeed is. In fact, $\varphi$ is a homomorphism with $\ker \varphi = \langle \rho^2 \rangle$, and $\bar{\varphi}$ is just the induced homomorphism from $\varphi$ that we obtain when we invoke the first isomorphism theorem. $\square$

The takeaway point is as follows. While $D_4$ and $\mathbb{Z}/2\mathbb{Z} \times \mathbb{Z} / 2\mathbb{Z}$ do not share the same relations, we can make some identifications in $D_4$ to recover the relations for the generators of $\mathbb{Z}/2\mathbb{Z} \times \mathbb{Z} / 2 \mathbb{Z}$. For example, $\rho$ can be “turned into” an element by making the identification $\rho^2 \sim 1$, as 4 is divisible by 2. It would not be possible to turn $\rho$ into an element of order 3, whence we cannot expect the mappping $\Gamma:D_4 \to \mathbb{Z}/3\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z}$ given by the same formula $\Gamma(\rho^n\tau^m) = (n,m)$ to be a group homomorphism. Indeed, $\rho^0 \tau^0= \rho^4\tau^0$, but

and so $\Gamma$ is not even well-defined.

This isn’t to say that the number of generators in the domain group should match up with the number of generators in the codomain group. Remember: generators do not behave like vector-spaces bases. It doesn’t even make sense to talk about the number of generators of a group. Moreover, we can just map multiple generators to one generator, or simply create a non-surjective map that misses some generators in the codomain group.

Example 3.5. The mapping $\varphi:\mathbb{Z} \times \mathbb{Z} \to \mathbb{Z}$ defined by the formula $\varphi((n,m)) = n$ is a surjective homomorphism. $\square$

Example 3.6. The mapping $\varphi:\mathbb{Z} \to \mathbb{Z} \times \mathbb{Z}$ defined by the formula $\varphi(n) = (n,0)$ is an injective homomorphism. $\square$

Here is a good rule of thumb for creating group homomorphisms via homomorphic extensions:

1. send generators to generators;
2. extend in a structure-preserving way (as described above several times);
3. make sure that the identity element gets sent to the identity element;
4. check that the function is well-defined.

Once you have many examples under your belt, you will be able to see intuitively when structure-preserving extensions work. We conclude the post with an application to the classification of cyclic groups.

Example 3.7 (Cyclic groups). Let $G$ be a cyclic group, so that $G = \langle x \rangle$ for some $x \in G$. We shall show that $G$ is isomorphic to either $\mathbb{Z}$ or $\mathbb{Z}/n\mathbb{Z}$ for some $n \in \mathbb{N}$; see Example 2.4.

We define a mapping $\varphi:\mathbb{Z} \to G$ by setting $\varphi(n) = x^n$. $\varphi$ is evidently a well-defined surjective function. Moreover,

$\varphi(n+m) = x^{n+m} = x^nx^m = \varphi(n)\varphi(m)$,

and so $\varphi$ is a homomorphism.

We now invoke the first isomorphism theorem to obtain the isomorphic relation $\varphi/\ker \varphi \cong G$. If $\ker \varphi$ is trivial, then $G \cong \mathbb{Z}$. If $\ker \varphi$ is nontrivial, then we determine $\ker \varphi$ as follows. Since $\ker \varphi$ is nontrivial, there exists at least one nonzero element of $\ker \varphi$. $\ker \varphi$ is closed under taking inverses, and so $\ker \varphi$ contains at least one positive integer. We let $n_0$ be the smallest positive integer in $\ker \varphi$.

We claim that $\ker \varphi = n_0 \mathbb{Z}$. If not, then, by long division, we can find a positive integer $n_1 \in \mathbb{Z}$ such that $n_1 = kn_0 + r$, where $r \in \mathbb{N}$ and $% $. Since $\ker \varphi$ is closed under taking inverses, $-kn_0$ is in $\ker \varphi$. Now, $\ker \varphi$ is closed under the group operation, and so $r = n_1 – kn_0$ is in $\ker \varphi$. This is evidently absurd, as $n_0$ was assumed to be the smallest positive integer. We conclude that $\ker \varphi = n_0 \mathbb{Z}$, whence $G \cong \mathbb{Z}/n_0\mathbb{Z}$. $\square$

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