19 MORE ON ISOMORPHISM

Note. If two finite groups are isomorphic, then they have the same order. Whence comes the simplest of all tests for showing that two groups are not isomorphic:

Theorem. If $G$ and $H$ are groups and $\abs{G}\ne\abs{H},$ then $G$ and $H$ are not isomorphic. Other properties that can be used to determine that two groups are not isomorphic are given next.

Theorem 19.1. Assume that $G$ and $H$ are groups and that $G\approx H.$

1. $\abs{G}=\abs{H}.$
2. If $G$ is Abelian, then $H$ is Abelian.
3. If $G$ is cyclic, then $H$ is cyclic.
4. If $G$ has a subgroup of order $n$ (for some positive integer $n$), then $H$ has a subgroup of order $n.$
5. If $G$ has an element of order $n,$ then $H$ has an element of order $n.$
6. If every element of $G$ is its own inverse, then every element of $H$ is its own inverse.
7. If every element of $G$ has finite order, then every element of $H$ has finite order.

Theorem 19.2. Isomorphism is an equivalence relation on the class of all groups.

Theorem 19.3. If $p$ is prime and $\abs{G}=p$ then group $G\approx\Z_p.$

Theorem. There is just one group of order $n$ iff $n$ is a prime or a product of distinct primes $p_1,\ldots,p_k$ such that $p_j\nmid(p_i-1)$ for $1\le i\le k$ and $1\le j\le k.$

Theorem. There are two isomorphism classes of groups of order $n=p^2:$ the group $\Z_{p^2}$ is in one class and $\Z_p\times\Z_p$ is in the other.

Theorem. There are five isomorphism classes of groups of order $n=p^3:$ three of these classes consist of Abelian groups (Example N 19.1): $\Z_{p^3},$ $\Z_{p^2}\times\Z_p,$ and $\Z_p\times\Z_p\times\Z_p;$ the other two classes consist of non-Abelian groups. See Problem 19.24 for the definition of direct products of more than two groups.

Fundamental Theorem of Finite Abelian Groups. If $G$ is a finite Abelian group, then $G$ is the direct product of cyclic groups of prime power order. Moreover, if $$G\approx A_1\times\cdots\times A_s$$ and $$G\approx B_1\times\cdots\times B_t,$$ where each $A_i$ and $B_j$ is cyclic of prime power order, then $s=t$ and, after suitable relabeling of subscripts, $\abs{A_i}=\abs{B_i}$ for $1\le i\le s.$

Example 19.1. Let $n=125=5^3.$ To apply the theorem, first determine all possible ways of factoring $125$ as a product of (not necessarily distinct) prime powers. Each factorization gives a different isomorphism class, so there are three isomorphism classes of Abelian groups of order 125. One represenative from each class is displayed.

Factorizations $(n=125)$	Isomorphism Class Representative
$5^3$	$\Z_{5^3}$
$5^2\cdot5$	$\Z_{5^2}\times\Z_{5}$
$5\cdot5\cdot5$	$\Z_{5}\times\Z_{5}\times\Z_{5}$

More generally,

Factorizations $(n=p^3)$	Isomorphism Class Representative
$p^3$	$\Z_{p^3}$
$p^2\cdot p$	$\Z_{p^2}\times\Z_{p}$
$p\cdot p\cdot p$	$\Z_{p}\times\Z_{p}\times\Z_{p}$

Example 19.2. Let $n=200=2^3\times5^2.$ There are six isomorphism classes of Abelian groups of this order. One representative from each class is displayed.

Factorizations $(n=200)$	Isomorphism Class Representative
$2^3\cdot5^2$	$\Z_{2^3}\times\Z_{5^2}$
$2^3\cdot5\cdot5$	$\Z_{2^3}\times\Z_5\times\Z_5$
$2^2\cdot2\cdot5^2$	$\Z_{2^2}\times\Z_2\times\Z_{5^2}$
$2^2\cdot2\cdot5\cdot5$	$\Z_{2^2}\times\Z_2\times\Z_5\times\Z_5$
$2\cdot2\cdot2\cdot5^2$	$\Z_2\times\Z_2\times\Z_2\times\Z_{5^2}$
$2\cdot2\cdot2\cdot5\cdot5$	$\Z_2\times\Z_2\times\Z_2\times\Z_5\times\Z_5$

Table 19.1. Number of isomorphism classes of groups of order $n$ for each $n$ from $1$ to $32.$

Order	Number of Groups
1	1
2	1
3	1
4	2
5	1
6	2
7	1
8	5
9	2
10	2
11	1
12	5
13	1
14	2
15	1
16	14
17	1
18	5
19	1
20	5
21	2
22	2
23	1
24	15
25	2
26	2
27	5
28	4
29	1
30	4
31	1
32	51

Problem 19.24. If $\{A_1,\ldots,A_n\}$ is any collection of groups, each with juxtaposition as operation, then the direct product of these groups is the group $A_1\times\cdots\times A_n$ $=\{(a_1,\ldots,a_n)\mid a_i\in A_i\}$ with operation $(a_1,\ldots,a_n) (b_1,\ldots,b_n)$ $=(a_1 b_1,\ldots,a_n b_n).$

Problem 19.25. An isomorphism of a group onto itself is called an automorphism. The set of all automorphisms of a group is itself a group with respect to composition. This group of automorphisms of a group is called the automorphism group of $G,$ and will be denoted $\Aut(G).$ Note that the elements of $\Aut(G)$ are mappings from $G$ onto $G.$