Functions between two groups

To compare two groups, we first talk about functions between two groups. A homomorphism between two groups is, roughly speaking, a function between them which preserves the (respective) group operations.

Definition 5.13.1   Let $G_1, G_2$ be groups, with $*_1$ denoting the group operation for $G_1$ and $*_2$ the group operation for $G_2$. A function $f : G_1 \rightarrow G_2$ is a homomorphism if and only if, for all $a,b \in G_1$, we have

$$f(a*_1 b) = f(a)*_2 f(b).$$

The subgroup $f(G_1)\ <\ G_2$ is called the image of $f$ and is sometimes denoted $im(f)$.

Example 5.13.2   Let $G$ be a group and $h$ a fixed element of $G$. Define $f : G \rightarrow G$ by

$$f(g) = h^{-1}*g*h, \ \ \ \ g \in G.$$

Then the following simple trick

$$f(a*b) = h^{-1}*(a*b)*h = h^{-1}*a*h*h^{-1}*b*h = f(a)*f(b)$$

shows that $f$ is a homomorphism. In this case, $im(f)=G$, i.e., $f$ is surjective.

Example 5.13.3   The function

$$sign: S_n \rightarrow \{\pm 1\},$$

which assigns to each permutation its sign, is a homomorphism, as was proven in chapter 3. Another reason why this is true is due to the fact that the sign of a permutation $g$ is the determinant of the associated permutation matrix $P(g)$. Since the determinant of the product is the product of the determinants (see Lemma 5.4.3), we have

$$sgn(gh)=\det P(gh)=\det (P(g)P(h))=\det P(g)\det P(h) =sign(g)sign(h),$$

for all $g,h\in S_n$. From this it follows that $sign$ is a homomorphism.

Lemma 5.13.4   If $f : G_1 \rightarrow G_2$ is a homomorphism then (a) $f(e_1)=e_2$, where $e_1$ denotes the identity element of $G_1$ and $e_2$ denotes the identity element of $G_2$, (b) $f(x^{-1})=f(x)^{-1}$, for all $x$ belonging to $G_1$, (c) $f(y^{-1}*_1 x*_1 y)=f(y)^{-1}*_2f(x)*_2 f(y)$, for all $x,y$ belonging to $G_1$, where $*_1$ denotes the group operation for $G_1$ and $*_2$ the group operation for $G_2$.

proof: (a) We have $f(x)=f(x*_1 e_1)=f(x)*_2 f(e_1)$, for any $x \in G_1$. Multiply both sides of this equation on the left by $f(x)^{-1}$. (b) We have, by part (a), $e_2=f(e_1)=f(x*_1 x^{-1})=f(x)*_2 f(x^{-1})$. Multiply both sides of this equation on the left by $f(x)^{-1}$. $\Box$

Definition 5.13.5   Let $G_1, G_2$ be finite groups. We say that $G_1$ embeds (or injects) into $G_2$ if there exists an injective homomorphism $f : G_1 \rightarrow G_2$. A homomorphism $f : G_1 \rightarrow G_2$ is a isomorphism if it is a bijection (as a function between sets). In this case, we call $G_1$ and $G_2$ isomorphic and write $G_1\cong G_2$. An isomorphism from a group $G$ to itself is called an automorphism.

The notion of an isomorphism is the notion we will use when we want to same two groups are ``essentially the same group'', i.e., one is basically a carbon copy of the other with the elements relabeled and the binary operation modified. For example, if you have any two groups of order $2$ then they must be isomorphic. (The interested reader can verify that the map which sends the identity of one group to the identity of the other and the only non-identity of one group to the only non-identity of the other must be a group isomorphism.) In other words, there is only one group of order $2$, up to isomorphism. Let $O(n)$ denote the the number of non-isomorphic groups of order $n$, so $O$ is a function $O:\mathbb{N}\rightarrow \mathbb{N}$ (where $\mathbb{N}=\{1,2,3,...\}$). This is a rather curiously behaving function.

Order	Group $G$	notes
2	$C_2$
3	$C_3$	$G\cong A_3$
4	$C_4$
4	$C_2\times C_2$	Klein 4-group
		$Aut(G)\cong GL(2,{\Bbb F}_2)$
5	$C_5$
6	$C_6=C_2\times C_3$
6	$S_3$	$Aut(G)=G$
		$G\cong GL(2,{\Bbb F}_2)$
7	$C_7$
8	$C_8$
8	$C_2\times C_4$
8	$C_2\times C_2\times C_2$	$Aut(G)\cong GL(3,{\Bbb F}_2)$
8	$D_4$
8	$Q$
9	$C_9$
9	$C_3\times C_3$	$Aut(G)\cong GL(2,{\Bbb F}_3)$
10	$C_{10}=C_2\times C_5$
10	$D_5$

Exercise 5.13.6   Pick a group at random from the above table and find a subgroup of the Rubik's cube group which is isomorphic to it. (This is not easy. See the chapter entitles ``Words which move'' in [J2] for some hints.)

Exercise 5.13.7   Prove the following: If $f : G_1 \rightarrow G_2$ is a homomorphism of groups then

$$f(G_1)=\{g\in G_2\ \vert\ g=f(x),\ {\rm for\ some}\ x\in G_1\}$$

is a subgroup of $G_2$.

Exercise 5.13.8   Find $O(n)$, $n=1,2,...,30$. (Hint: Use GAP or MAGMA. This is practically impossible to do ``by hand''.)

Exercise 5.13.9   Let $G$ be the group $G=\{f_1, f_2, f_3, f_4, f_5, f_6\}$ as given in Exercise 5.3.7. Is $G$ isomorphic to $S_3$? To $C_2 \times C_3$? Explain.

Exercise 5.13.10   Let $G$ be any group and let $h \in G$ be any fixed element. Let $\phi:G\rightarrow G$ be defined by $\phi(g)=h^{-1}gh$, for $g\in G$. Show that $\phi$ is an isomorphism.

Exercise 5.13.11   Let

$$G=\{ \left( \begin{array}{cc} 1 &0 \\ 0 &1 \end{array... ...begin{array}{cc} -1 &0 \\ 0 &-1 \end{array} \right) \}$$

and $H=\mathbb{Z}/2\mathbb{Z}\times \mathbb{Z}/2\mathbb{Z}$. Show that $G\cong H$.

Exercise 5.13.12   Show the following: (a) If $m\geq 1$ then $C_m\cong \mathbb{Z}/m\mathbb{Z}$. (b) If $m>1,n>1$ have no prime divisors in common, i.e., if $m,n$ are relatively prime, then $C_m\times C_n\cong C_{mn}$. (Hint: (a) Let $a\in C_m$ be a generator. The map $f=f_{m,a}:C_m\rightarrow \mathbb{Z}/m\mathbb{Z}$ which satisfies $f(a)=1$ extends to a unique homomorphism between these groups. Show that this is an isomorphism. (b) First, show $g(x+mn\mathbb{Z})=(x+m\mathbb{Z},x+n\mathbb{Z})$, for all $x\in \mathbb{Z}$ defines a homomorphism $\mathbb{Z}/mn\mathbb{Z} \rightarrow \mathbb{Z}/m\mathbb{Z}\times \mathbb{Z}/n\mathbb{Z}$. Next, using the assumption that $m,n$ are relatively prime, show that this function $g$ is an injection. Conclude, since the domain and range of $g$ both have the same cardinality, that $g$ is also a surjection. Finally, using part (a), show $C_m\times C_n\cong C_{mn}$.)

Exercise 5.13.13   Show part (c) in Lemma 5.13.4. (Hint: use the definition of homomorphism and part (b)).

Exercise 5.13.14   Let

$$A= \left( \begin{array}{ccc} 0 & 1 & 0\\ 1 & 0 & 0\\ ... ... 1 & 0 & 0\\ 0 & 0 & 1\\ 0 & 1 & 0 \end{array} \right)$$

Now, let $G = \langle A,B\rangle$ denote the group of all matrices which can be written as any arbitrary product of these two matrices (in any order and with as many terms as you want). We have

$$\begin{array}{c} G=\{ I_3= \left( \begin{array}{ccc} 1 ... ... & 1 & 0\\ 1 & 0 & 0 \end{array} \right) \} \end{array}$$

(The interested reader may want to try to check this by regarding each such matrix as a permutation matrix.) Define $f : G \rightarrow S_3$ by

$g$	$f(g)$
$I_3$	$1$
$A$	$s_1$
$B$	$s_2$
$A*B$	$s_1*s_2$
$B*A$	$s_2*s_1$
$A*B*A$	$s_1*s_2*s_1$

Show that this is a homomorphism.