**7.**
Find all abelian groups of order 108 (up to isomorphism).

*Solution:*
The prime factorization is
108 = 2^{2} **·** 3^{3}.
There are two possible groups of order 4:
**Z**_{4} and
**Z**_{2}
×
**Z**_{2} .
There are three possible groups of order 27:
**Z**_{27} ,
**Z**_{9}
×
**Z**_{3} ,
and
**Z**_{3}
×
**Z**_{3}
×
**Z**_{3} .
This gives us the following possible groups:

**Z**_{4}×**Z**_{27}**Z**_{2}×**Z**_{2}×**Z**_{27}**Z**_{4}×**Z**_{9}×**Z**_{3}**Z**_{2}×**Z**_{2}×**Z**_{9}×**Z**_{3}**Z**_{4}×**Z**_{3}×**Z**_{3}×**Z**_{3}**Z**_{2}×**Z**_{2}×**Z**_{3}×**Z**_{3}×**Z**_{3}.

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**8.**
Let G and H be finite abelian groups,
and assume that G × G
is isomorphic to H × H.
Prove that G is isomorphic to H.

*Solution:*
Let p be a prime divisor of |G|,
and let q = p^{m}
be the order of a cyclic component of G.
If G has k such components,
then G × G
has 2k components of order q.
An isomorphism between G × G
and H × H
must preserve these components,
so it follows that H also has k cyclic components of order q.
Since this is true for every such q,
it follows that G H.

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**9.**
Let G be an abelian group which has 8 elements of order 3,
18 elements of order 9, and no other elements besides the identity.
Find (with proof) the decomposition of G
as a direct product of cyclic groups.

*Solution:*
We have |G| = 27.
First, G is not cyclic since there is no element of order 27.
Since there are elements of order 9,
G must have **Z**_{9} as a factor.
To give a total of 27 elements,
the only possibility is
G
**Z**_{9}
×
**Z**_{3}.

Check: The elements 3 and 6 have order 3 in **Z**_{9},
while 1 and 2 have order 3 in **Z**_{3}.
Thus the following 8 elements have order 3 in the direct product:
(3,0),
(6,0),
(3,1),
(6,1),
(3,2),
(6,2),
(0,1), and
(0,2).

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**10.**
Let G be a finite abelian group such that |G| = 216.
If | 6 G | = 6, determine G up to isomorphism.

*Solution:*
We have 216 = 2^{3} **·** 3^{3},
and 6G
**Z**_{2}
×
**Z**_{3} since it has order 6..
Let H be the Sylow 2-subgroup of G,
which must have 8 elements.
Then multiplication by 3 defines an automorphism of H,
so we only need to consider 2H.
Since 2H **Z**_{2},
we know that there are elements not of order 2,
and that H is not cyclic, since
2 **Z**_{8}
**Z**_{4}.
We conclude that
H
**Z**_{4}
×
**Z**_{2}.

A similar argument shows that the Sylow 3-subgroup K of G,
which has 27 elements, must be isomorphic to
**Z**_{9}
×
**Z**_{3}.

Using the decomposition in Theorem 7.5.4,
we see that

G
**Z**_{4}
×
**Z**_{2}
×
**Z**_{9}
×
**Z**_{3}.

(If you prefer the form of the decomposition in
Proposition 7.5.5,
you can also give the answer in the form

G
**Z**_{36}
×
**Z**_{6}.)

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**11.**
Apply both structure theorems
to give the two decompositions of the abelian group
**Z**_{216}^{×}

*Solution:*
**Z**_{216}^{×}
**Z**_{8}^{×}
×
**Z**_{27}^{×}
**Z**_{2}
×
**Z**_{2}
×
**Z**_{27}^{×}

Since 27 is a power of an odd prime,
it follows from Corollary 7.5.11
that **Z**_{27}^{×} is cyclic.

This can also be shown directly
by guessing that 2 is a generator.

Since **Z**_{27}^{×}
has order 3^{3} - 3^{2} = 18,
an element can only have order 1, 2, 3, 6, 9 or 18.
We have

2^{2} = 4,

2^{3} = 8,

2^{6} 8^{2}
10,
and

2^{9}
2^{3} **·**
2^{6}
8 **·** 10 -1,

so it follows that 2 must be a generator.

We conclude that
**Z**_{216}^{×}
**Z**_{2}
×
**Z**_{2}
×
**Z**_{18}.

To give the first decomposition, recall that
Theorem 7.5.4
states that any finite abelian group is isomorphic to a direct product
of cyclic groups of prime power order.
In this decomposition we need to split
**Z**_{18} up into cyclic subgrops of prime power order,
so we finally get the decomposition

**Z**_{216}^{×}
**Z**_{2}
×
**Z**_{2}
×
**Z**_{2}
×
**Z**_{9}.

On the other hand,
Proposition 7.5.5
describes the second decomposition,
where any finite abelian group is written as a direct product
of cyclic groups in which the orders any component
is a divisor of the previous one.
To do this we need to group together
the largest prime powers that we can.
In the first decomposition,
we can combine **Z**_{2} and **Z**_{9} to get
**Z**_{18} as the first component.
We end up with

**Z**_{216}^{×}
**Z**_{18}
×
**Z**_{2}
×
**Z**_{2}

as the second way of breaking
**Z**_{216}^{×}
up into a direct product of cyclic subgroups.

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**12.**
Let G and H be finite abelian groups,
and assume that they have the following property.
For each positive integer m,
G and H have the same number of elements of order m.
Prove that G and H are isomorphic.

*Solution:*
We give a proof by induction on the order of |G|.
The statement is clearly true for groups of order 2 and 3,
so suppose that G and H are given,
and the statement holds for all groups of lower order.
Let p be a prime divisor of |G|,
and let G_{p} and H_{p} be the Sylow p-subgroups of G and H,
respectively.
Since the Sylow subgroups contain all elements of order a power of p,
the induction hypothesis applies to G_{p} and H_{p}.
If we can show that
G_{p}
H_{p} for all p,
then it will follow that G H,
since G and H are direct products of their Sylow subgroups.

Let x be an element of G_{p} with maximal order
q = p^{m}.
Then < x > is a direct factor of G_{p} by Lemma~7.5.3,
so there is a subgroup G' with
G_{p} = < x > × G'.
By the same argument we can write
H_{p} = < y > × H',
where y has the same order as x.

Now consider < x^{p} >
× G' and
< y^{p} > × H'.
To construct each of these subgroups we have removed
elements of the form (x^{k},g'),
where x^{k} has order q and g' is any element of G'.
Because x has maximal order in a p-group,
in each case the order of g' is a divisor of q,
and so (x^{k},g') has order q since
the order of an element in a direct product
is the least common multiple of the orders of the components.
Thus to construct each of these subgroups we have removed
(p^{m} - p^{m-1}) **·** |G'| elements,
each having order q.
It follows from the hypothesis
that we are left with the same number of elements of each order,
and so the induction hypothesis implies that
< x^{p} > × G'
and < y^{p} > × H'
are isomorphic.
But then G' H', and so
G_{p} H_{p},
completing the proof.

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