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**Definition 3.4.1.**
Let G_{1} and G_{2} be groups,
and let µ : G_{1} -> G_{2}
be a function. Then µ is said to be a
**group isomorphism**
if

**(i)**µ is one-to-one and onto and**(ii)**µ (ab) = µ (a) µ (b) for all a,b in G_{1}.

**(a)**The inverse of a group isomorphism is a group isomorphism.**(b)**The composition of two group isomorphisms is a group isomorphism.

**Proposition 3.4.3.**
Let
µ : G_{1} -> G_{2}
be an isomorphism of groups.

**(a)**If a has order n in G_{1}, then µ (a) has order n in G_{2}.**(b)**If G_{1}is abelian, then so is G_{2}.**(c)**If G_{1}is cyclic, then so is G_{2}.

**Proposition 3.4.4.**
Let G_{1} and G_{2} be groups, and let
µ : G_{1} -> G_{2}
be a function such that

µ (ab) = µ (a) µ (b)

for all a,b in G

**Proposition 3.4.5.**
If m,n are positive integers such that gcd(m,n)=1, then

**Z**_{m} × **Z**_{n}
**Z**_{mn}.

µ (ab) = µ (a) µ (b),
for all a,b in G_{1}

In terms of the respective group multiplication tables
for G_{1} and G_{2},
the existence of an isomorphism guarantees that there is a way
to set up a correspondence between the elements of the groups
in such a way that the group multiplication tables
will look exactly the same.

From an algebraic perspective,
we should think of isomorphic groups as being essentially the same.
The problem of finding all abelian groups of order 8
is impossible to solve, because there are infinitely many possibilities.
But if we ask for a list of abelian groups of order 8
that comes with a guarantee that *any* possible abelian group
of order 8 must be isomorphic to one of the groups on the list,
then the question becomes manageable.
In fact, we can show (in Section 7.5)
that the answer to this particular question is the list

**Z**_{8},
**Z**_{4} × **Z**_{2},
**Z**_{2} × **Z**_{2} × **Z**_{2}.

To show that two groups G_{1} and G_{2} are isomorphic,
you should actually produce an isomorphism
µ : G_{1} -> G_{2}.
To decide on the function to use,
you probably need to do some experimentation
(compute some products)
in order to understand why the group operations are similar.

In some ways it is harder to show that two groups are *not* isomorphic.
If you can show that one group has a property
that the other one does not have,
then you can decide that two groups are not isomorphic
(provided that the property would have been transferred by any isomorphism).
Suppose that G_{1} and G_{2} are isomorphic groups.
If G_{1} is abelian, then so is G_{2};
if G_{1} is cyclic, then so is G_{2}.
Furthermore, for each positive integer n,
the two groups must have exactly the same number of elements of order n.
Each time you meet a new property of groups,
you should ask whether it is preserved by any isomorphism.

**21.**
Show that **Z**_{17}^{×}
is isomorphic
to **Z**_{16}.
*Solution*

**22.**
Let µ : **R**^{×} -> **R**^{×}
be defined by
µ (x) = x^{3}, for all x in **R**.
Show that µ is a group isomorphism.
*Solution*

**23.**
Let G_{1}, G_{2}, H_{1}, H_{2} be groups,
and suppose that
µ_{1} : G_{1} -> H_{1} and
µ_{2} : G_{2} -> H_{2} are group isomorphisms.
Define

µ : G_{1} × G_{2} ->
H_{1} × H_{2}

µ (x_{1},x_{2})
= ( µ_{1} (x_{1}),
µ_{2} (x_{2})),

**24.**
Prove that the group
**Z**_{7}^{×} ×
**Z**_{11}^{×}
is isomorphic
to the group **Z**_{6} × **Z**_{10}.
*Solution*

**25.**
Define µ : **Z**_{30} × **Z**_{2}
-> **Z**_{10} × **Z**_{6}
by

µ ( [n]_{30} , [m]_{2} )
= ( [n]_{10} , [4n+3m]_{6} ),

**26.**
Let G be a group, and let H be a subgroup of G.
Prove that if a is any element of G, then the subset

aHa^{-1}
= { g in G | g = aha^{-1} for some h in H }

**27.**
Let G, G_{1}, G_{2} be groups.
Prove that if G is isomorphic
to G_{1} × G_{2},
then there are subgroups H and K in G such that

(i) H K = { e },

(ii) HK = G, and

(iii) hk=kh for all h in H and k in K.
*Solution*

**28.**
Show that for any prime number p,
the subgroup of diagonal matrices in
GL_{2} (**Z**_{p} )
is isomorphic
to **Z**_{p}^{×} × **Z**_{p}^{×}.
*Solution*

**29.**
(a) In the group G = GL_{2} (**R**)
of invertible 2 × 2 matrices with real entries, show that

H =

is a subgroup of G.(b) Show that H is isomorphic to the group

**30.**
Let G be the subgroup of GL_{2} (**R**) defined by

G = .

Show that G is not isomorphic to the direct product
**31.**
Let H be the following subgroup of the group
G = GL_{2} (**Z**_{3}).

H =

Show that H is isomorphic to the symmetric group S
**32.**
Let G be a group, and let S be any set
for which there exists a one-to-one and onto function
µ : G -> S.
Define an operation on S by setting
x_{1} · x_{2}
= µ ( µ ^{-1} (x_{1})
µ ^{-1} (x_{2}) ),
for all x_{1}, x_{2} in S.
Prove that S is a group under this operation,
and that µ is actually a group isomorphism.
*Solution*

**Lab 1.**
The aim of this problem is to show that
Table 3.3.2 in the text
(see page 103 of the text or the
online table)
and the table shown by
** Groups15**
for the cyclic group Z

**Lab 2.**
The aim of this problem is to see whether
Table 3.3.3 in the text
(see page 104 of the text or the
online table)
and the table shown by
** Groups15**
for the group S

The previous problem is really a specific
application of Exercise 3.3.14 in the text,
which asks you to prove that any group of order 6
is either cyclic or isomorphic to S_{3}.

Corollary 3.2.12 shows that any group of prime order is cyclic.
The analysis of groups of order 4 given on pages 103 and 104 of the text
shows that if |G| = 4, then either G has at least one element of order 4,
and is therefore cyclic, or G has three elements of order 2
(not counting the identity).
In the latter case G must be isomorphic to
Z_{2} × Z_{2}.
These comments give you enough information
to do a deeper investigation of the lab questions in Section 3.3.

**Lab 3.**
Show that the group
Z_{3} Z_{4}
(of order 12)
in
** Groups15**
is the internal semidirect product of a normal subgroup of order 3
and a cyclic subgroup of order 4.
(See the lab problem 3.3.10.)

**Lab 4.**
Show that the group
D_{6} (of order 12) in
** Groups15**
is the internal semidirect product of a normal subgroup
isomorphic to Z

**Lab 5.**
Show that the group
A_{4} (of order 12) in
** Groups15**
is the internal semidirect product of a normal subgroup
isomorphic to Z

**Lab 6.**
Find an isomorphism between the group Z_{12},
as described in
** Groups15**,
and the group Z

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