136.330 Intro to Algebra
Test 3 Solutions
(November 27, 2002)
1. Suppose A and
B are two groups such that 
and 
.
(a) How many subgroups of 
of order 6 are
there? (Do not forget to justify your answer.)
(b) Show that 
is a subgroup of
.
(c) What is the order of the
element 
in 
, where 
and 
?
Solution. (a) The
order of 
is (4)(5)=20.
The order of each subgroup must divide 20 (Lagrange). So, there are no
subgroups of order 6.
(b) 
, so we have closure under the operation in 
. Further, 
so the set 
is also closed
under taking inverses. So, it is a subgroup of 
.
(c) The order of 
is obviously the
same as the order of x, and the latter is 5 since 
and since 
is a prime
number.
2. Show that a group
G of order 27 is either cyclic or is such that 
for all g in G.
Solution. Suppose G is not cyclic. Then no element in G has the order of 27. Since the order of any element divides 27, the elements of G are of order 1, 3, or 9. In all of these cases, the ninth power of that element is e.
3. The mapping 
is defined by 
(Please note
that we are using the additive notation here; so, for example 
is short for 
, k many times).
(a) Find 
.
(b) Show that 
is a
homomorphism.
(c) Find the kernel of 
.
Solution. (a) 
.
(b) 
and 
. Since 
(by definition
of addition in 
) it follows that f is a homomorphism.
(c) The kernel of f is the set of
elements of 
that are mapped
to the identity in 
. Obviously, the kernel of f in this example is 
.
4. (a) How many isomorphisms 
? Find all of them.
(b) Show that 
is not isomorphic to 
.
Solution.
(a) Since homomorphism map the identity
of a group to the identity of the image group, so do isomorphisms; so for any
isomorphism from 
to 
, we have 
. Since isomorphisms are bijections, there is only one
remaining choice for the image of 
, namely 
. So, the only isomorphism is the identity isomorphism.
(b) 
is a cyclic
group. We show that 
is not cyclic
and the claim in the problem follows.
The non-cyclicity of
is a consequence of the observation that no element in 
generates 
(it is easy to see that the order of every non-identity element
in 
is 2).
5. (a) State the definition of a normal subgroup H of a group G.
(b) Consider the subgroup 
of the group of
permutations
. Show that H is a normal subgroup of 
. (Just show it is normal; you do not have to show it is a
subgroup.)
Solution. (b) Notice
that H is the subgroup of 
made of all even
permutation. Then notice that for every permutation 
and every even
permutation 
, 
must also be
even. So for every permutation 
in 
, 
is again the set
of all even permutations, and so 
. Thereby H is a
normal subgroup of 
.
[A walking proof is also acceptable.]