Austin Mohr

Math 701
Homework 5

Problem 17
Prove that if G, H, and K are finite Abelian groups and G H
= G K, then H
= K.
Proof. Since G, H, and K are finite Abelian groups, they are each isomorphic to some direct product of
cyclic groups of prime power order.
Observe that, since G H = G K and all the groups are finite,
|G||H| = |G H| = |G K| = |G||K|
Hence, |H| = |K|.
Now, suppose for the sake of contradiction that H is not isomorphic to K. Since they are of the same
finite order, it must be that one of the cyclic groups comprising H appears with different multiplicity than
in K (note one of these multiplicites may be 0). Let the order of this cyclic factor be pk for some prime p.
We see that H and K necessarily have differing numbers of elements of order pk . This implies that G H
and G K do not have the same number of elements of order pk , and so are not isomorphic, which is a
contradiction. Therefore, it must be that H = K.
Problem 18
Prove that every group of order 35 is cyclic.
Proof. Observe first that 35 = 5 7. Consider the Sylow p-subgroups of G. Sylows theorem gives
n5 1 (mod 5) and n5 | 7, so n5 = 1
n7 1 (mod 7) and n7 | 5, so n7 = 1

Hence, G has a unique Sylow 5-subgroup N5 and a unique Sylow 7-subgroup N7 . The uniqueness of each
of these groups implies that they are normal in G.
Observe next that N5 N7 is trivial, since only the identity element can have order dividing both |N5 |
and |N7 |. By the Third Isomorphism theorem, we have
N5 N7 N5
N7 = N5 N7 = N5
Applying Lagranges theorem, we have
 
N5 N7
N7  = |N5 |


|N5 N7 |
= |N5 |
|N7 |
|N5 N7 | = |N5 ||N7 | = |G|

Hence, G = N5 N7 .
Now, both of N5 and N7 are cyclic (and so Abelian), since they are of prime order. We claim that G is
also Abelian. Let (a1 , b1 ) and (a2 , b2 ) be elements of G. It follows that

(a1 , b1 ) (a2 , b2 ) = (a1 a2 , b1 b2 )

= (a2 a1 , b2 b1 )
= (a2 , b2 )(a1 , b1 )

Hence, G is Abelian. We see that G is a finite Abelian group with k 2 not dividing its order for all k > 1.
By problem 20, we conclude that G is cyclic.

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Problem 19
Describe, up to isomorphism, all groups of order 1225.
Proof. Observe first that 1225 = 52 72 . Consider the Sylow p-subgroups of G. Sylows theorem gives
n5 1 (mod 5) and n5 | 49, so n5 = 1

n7 1 (mod 7) and n7 | 25, so n7 = 1

Hence, G has a unique Sylow 5-subgroup N5 and a unique Sylow 7-subgroup N7 . The uniqueness of each
of these groups implies that they are normal in G.
Observe next that N5 N7 is trivial, since only the identity element can have order dividing both |N5 |
and |N7 |. By the Third Isomorphism theorem, we have
N5 N7 N5
N7 = N5 N7 = N5
Applying Lagranges theorem, we have
 
N5 N7
N7  = |N5 |


|N5 N7 |
= |N5 |
|N7 |
|N5 N7 | = |N5 ||N7 | = |G|

Hence, G = N5 N7 .
We proceed by showing N5 is Abelian. Since N5 is of prime power order, it has a nontrivial center
Z(N5 ). Furthermore, Z(N5 ) is normal in N5 , so N5Z(N ) is a group of size 1 or 5. If it is of size 5, then
5
it is cyclic. We claim that this is impossible in general.
Claim 1. If a group G properly contains its center, then GZ(G) is not cyclic.

Proof. Suppose, to the contrary, that GZ(G) is cyclic generated by aZ(G). We argue that G is Abelian.
Let b and c be elements of G. We can find integers m and n so that

bZ(G) = am Z(G)
cZ(G) = an Z(G)

This further implies that we can find elements d and e in Z(G) so that

b = am d
c = an e

Observe that d and e commute freely with any element since they are in the center. Furthermore, powers of
a commute with each other. It follows that

bc = (am d)(an e)
= (an e)(am d)
= cb

Hence, G is Abelian, so Z(G) = G, which contradicts our assumption that G properly contains its center.
Therefore, we conclude that GZ(G) is not cyclic.

Citing the claim above, we conclude that N5Z(N ) is of size 1. In other words, N5 is Abelian.
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Similarly, we can show that N7 is Abelian (replace every occurrence of 5 with 7 in the argument for
N5 ).

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 Applying the Fundamental Theorem of Finite Abelian Groups, we conclude that G is isomorphic to one
of
Z49 Z25
Z49 Z5 Z5
Z25 Z7 Z7
Z7 Z7 Z5 Z5

Problem 20
Let G be a finite Abelian group. Prove that if |G| is not divisible by k 2 for any k > 1, then G is cyclic.

Proof. By the Fundamental Theorem of Finite Abelian Groups, G has a unique decomposition as a direct
product of cyclic groups of prime power order. More precisely,
G
= Zp1 n1 Zpk nk
where each pi is a prime number and each ni 1. Since k 2 does not divide the order of G for any k > 1, it
must be that ni = 1 for all i (otherwise, p2i divides the order of G for some i). We have now
G
= Zp1 Zpk
We see that |G| = p1 pk . We claim that G has an element of order p1 pk , and hence is cyclic.
Consider the element (1, . . . , 1) of G. We see that (1, , 1)m = (0, , 0) if and only if m is the least
| {z }
k times
common multiple of p1 , , pk . Since the pi are prime (and so relatively prime), m = p1 pk . That is,
(1, , 1) has order p1 pk , and so G is cyclic.