This exercise classifies the groups of order 60. (There are thirteen isomorphism types.)

- If is not normal, then contains at least two Sylow 5-subgroups. By Proposition 4.21, is simple. By Proposition 4.23, .
- Suppose is normal in and is not normal in . Since is normal in , is a subgroup. Moreover, since by Lagrange, ; there is only one group of order 15 up to isomorphism, so that . Note that has a unique (hence normal) subgroup of order 3- namely . Since is normal, we have . Thus 15 divides . Now ; thus 15 also divides . By Cauchy, contains an element of order 5, and thus contains . Then . Now recall that for some integer . By Sylow’s Theorem, , and in fact, . Let , , , and be the Sylow 3-subgroups of .
Recall that and is normal, so that and . Now for all ; since each contains a unique Sylow 3-subgroup, if then , a contradiction when . Thus for all distinct and , so that . Note that each contains elements of order 15 (where denotes the Euler totient). Moreover, if , then and for some , so that . Counting elements, we have . Thus every element of order 15 in is in some , no element is in more than one , and each contains precisely 8 elements of order 15. Thus has exactly elements of order 15.

Now let . Recall that for some . Now , since is normal in . By Sylow’s Theorem, all the are conjugate to . Thus we have for all . In particular, contains at least 32 elements; by Lagrange, , and we have .

The Sylow 3-subgroups of intersect trivially, so that every element of order 3 is in exactly one Sylow 3-subgroup. Each such subgroup contains 2 elements of order 3; thus contains exactly elements of order 3. Let denote the set of all elements of order 3 or 15 in ; note that , so that . Now let be a Sylow 2-subgroup of . Since is normal, is a subgroup. Since , . No element of has order 3 or 15 by Lagrange, so that . Counting elements, we see that . Moreover, if is a subgroup of order 20, then , so that . Thus, is the unique subgroup of order 20 in ; thus is characteristic, hence normal, in . Note for future reference that every element of has order 1, 2, 3, 4, 5, 10, or 15.

Now every Sylow 2-subgroup of is contained in , and in fact since the Sylow 2-subgroups of both groups have order 4. By Sylow’s Theorem, . Recall that . Since , we have , where divides . Thus , so that contains an element of order 3 by Cauchy. Now some Sylow 3-subgroup of is contained in ; without loss of generality, say . Now , since , so that is normal. Thus is normal. Note that every element of has order 1, 2, 3, or 4; there is one identity, and there are 3 elements of order 2 or 4; the remaining elements have order 3. Thus for all . Now let . Now for each , and is normal in , so that is a subgroup. Counting elements we see that for each . Now let ; we have . Thus is normal. Finally, note that and is a group of order 12; every group of order 12 has a unique Sylow 3-subgroup except A_4$. Thus .

Finally, since are normal and intersect trivially, we have . That is, .

- Before we proceed, we prove some technical lemmas.
Lemma 1: Fix integers such that, for all positive integers , (mod 5). Then (mod 5) and (mod 2). Proof: Suppose (mod 5), and let . Then (mod 5), and since is invertible mod 5, (mod 5)$ for all positive integers . This is absurd; consider . Thus (mod 5). Now let (mod 5); we have 4^n \equiv 1$ (mod 5); since (mod 5), we have (mod 2).

Lemma 2: Fix integers such that, for all positive integers , (mod 3). Then (mod 3) and (mod 2). Proof: Suppose (mod 3), and let . Then , so (mod 3) for all ; this is absurd- consider . Thus (mod 3). Now we have (mod 3); let (mod 3). Then (mod 3). Thus (mod 2).

Lemma 3: Fix integers such that, for all positive integers , (mod 5). Then (mod 5) and (mod 4). Proof: Suppose (mod 5), and let . Then (mod 5), so that (mod 5) for all positive ; this is absurd (consider ) so that (mod 5). Thus (mod 5). If (mod 5), we have (mod 5). Thus (mod 4).

Suppose now that are normal and let be a Sylow 2-subgroup of . As above, , and is normal. Moreover, . By the recognition theorem for semidirect products, , where . Evidently, classifying the groups of order 60 which have normal Sylow 5- and Sylow 3-subgroups is equivalent to describing the possible semidirect products , where . We carry out this classification now. Let .

Note that , where and . Moreover, being cyclic, every subgroup of is characteristic. By this previous exercise, we have . Write , where , , , and .

Suppose now that . There is a unique group homomorphism for each element of order dividing 4 in . Every element of has order dividing 4, so there are 8 distinct homomorphisms.

- If , then .
- If , then is indeed a nonabelian group of order 60. Note that if , then , where and . By a lemma to a previous exercise, . Moreover, note that .
- If , then is a nonabelian group of order 60. Note that if , then , where and . By a lemma to a previous exercise, . Moreover, note that .
- If , then is a nonabelian group of order 60. Moreover, note that .
- If , then is a nonabelian group of order 60. Moreover, note that .
- If , then is a nonabelian group of order 60. Moreover, note that .

We claim that these six groups are pairwise distinct.

- Note that is abelian and thus distinct from the remaining five.
- Note that and are distinct from , , and because and have trivial kernels while , , and do not. We now find presentations for and .
is generated by , , and , and . Now and . Thus has the presentation .

is generated by , , and , and . Now and . Thus has the presentation .

Now we compute the centers of and .

Every element of can be written in the form , where , , and . There are 60 such expressions, and we know by other means that . Thus every element can be written in this way uniquely. Let and let . Then, using a lemma to this previous exercise, is equal to . Comparing exponents, we have (mod 5). By Lemma 2, we have (mod 5) and (mod 4). Thus . Clearly , so that .

Similarly, every element of can be written uniquely in the form , where , , and . Let and let . Then is equal to . Comparing exponents, we have (mod 5) and (mod 3). By Lemmas 2 and 3, we have (mod 5), (mod 3), and (mod 4). Thus , and we have .

In particular, since their centers are not isomorphic.

- Next, we find presentations for , , and .
is generated by , , and , and . Now . Similarly, . Thus has the presentation .

is generated by , , and , and . Now and . Thus has the presentation .

is generated by , , and , and . Now and . Thus has the presentation .

Now we compute the centers of , , and .

Note that every element of can be written in the form , where , , and . There are 60 such expressions, and we know by other means that . Thus this representation of an element in is unique. Now let , and let . Then, using a lemma to this previous exercise, is equal to . Comparing exponents, we have (mod 5). By Lemma 1, this implies that (mod 5) and (mod 2). Thus or . Hence . Moreover, it is clear that and since . Thus .

Similarly, every element of can be written uniquely in the form . Let and let . Then is equal to . Comparing exponents, we have (mod 3) and (mod 2) by Lemma 2. Thus or , so that . We can see that and since . Thus .

Similarly, every element of can be written uniquely in the form . Let and let . Then is equal to . Comparing exponents, we have (mod 5) and (mod 3). By Lemmas 1 and 2, we have (mod 5), (mod 3), and (mod 2). Thus . Moreover, note that and , so that . Thus .

In particular, , , and are mutually nonisomorphic because their centers are not isomorphic.

Thus there are precisely 6 nonisomorphic groups of order 60 which have normal Sylow 5- and Sylow 3-subgroups and whose Sylow 2-subgroups are cyclic.

Suppose now that . Every homomorphism is uniquely determined by and , and such an assignment determines a homomorphism precisely when and divide 2. We can easily see that the number of elements of order dividing 2 in is four: , , , and . This gives us 16 distinct homomorphisms . We summarize these mappings in the following table; rows and columns are indexed by and , respectively.

Also, define the following mappings .

- and
- and
- and
- and
- and

Note the following.

- and
- and
- and
- , , , , and

It is easy to see that each is an isomorphism; thus, by a lemma to a previous exercise, the semidirect products formed by any two which appear in an equation are isomorphic. This gives us at most five distinct groups, which we name as follows.

We claim that these five are indeed distinct.

Note that is abelian, while the other four are not; thus is distinct from the other candidates.

We can easily see that while ; thus is distinct from , , and .

Next we compute presentations for , , and .

is generated by , , , and , and we have and . Now , , , and . Thus has the presentation , , , .

is generated by , , , and , and we have and . Now , , , and . Thus has the presentation .

is generated by , , , and , and we have and . By a similar process, we see that has the presentation .

Next, we compute the centers of , , and .

Every element of can be written in the form , where , , and . There are 60 such forms, and we know by other means that . Thus every element can be written in this form uniquely. Let and let . Now is equal to . Comparing exponents, we have (mod 5). By Lemma 1, this implies (mod 5) and (mod 2). Thus . We can easily see that , so that .

Every element of can also be written uniquely in the form , where , , and . Let and let . Now is equal to . Comparing exponents and using Lemma 2, we see that (mod 3) and (mod 2). Thus . Clearly , so that .

Every element of can also be written uniquely in the form , where , , and . Let and let . Now is equal to . Comparing exponents and using Lemma 2, we see that (mod 5), (mod 3), and (mod 2). Thus . Clearly , so that .

Clearly then , , and are distinct because their centers are not isomorphic.

This completes the classification of groups of order 60.