Math 546
Problem Set #13
1. Show that if G is a finite group and H is a non-empty subset of G, then H is a
subgroup of G if and only of H is closed under the operation on G.
Hint: For a given a in G, the elements:
a,!a2,!a3,!a4,…
cannot all be different.
Solution: We need to verify that e belongs to H and that for every element of H
its inverse also belongs to H.
We can accomplish both tasks essentially at the same time. Suppose that
a!H
.
Then the elements:
a,!a2,!a3,!a4,…
cannot all be different.
Hence there must exist some integers
0<k<n
such that
an=ak
. But then by the
power laws for groups, let
t=n!k
, then
at=an!k=a0=e
. Thus since H is
closed, e must belong to H.
Also, we have
at!1a=at=e
and so
at!1
is the inverse of a and belongs to H.
Note that this is a very useful result – for finite groups it is easy to check that a
subset is a subgroup – just check that it is closed under the operation.
2. Give an example of an infinite group G and a subset of G that is closed under the
group operation but is not a subgroup.
Solution: Take the integers under ordinary addition and let H be the set of
positive integers. H is closed under ordinary addition, but it is not a subgroup.
3. Verify that if p is a prime, then
!
(p)=p"1
.
Show that if p and q are primes, then
!
(pq)=(p"1)(q"1)
.
Solution: The numbers between 1 and
pq !1
that are not relatively prime to pq
are
p,!2p,!3p,…,!(q!1)p, and q,!2q,…,!(p!1)q
. So there are
p!1+q!1=p+q!2
numbers from 1 to
pq !1
that are not relatively prime to
pq. So the number of integers from 1 to
pq !1
that are relatively prime to pq is
pq !1
( )
!p+q!2
( )
=pq !p!q+1=p!1
( )
q!1
( )
.
Some other results that we won’t prove here, but you may take as known, are:
(i). If
gcd(n,m)=1
, then
!
(nm)=
!
(n)
!
(m)
.
(ii).If p is a prime, then
!
(pn)=pn"1(p"1)
.
4. Prove: If a and n are positive, relatively prime integers, then
a
!
(n)"1 mod n
.
This result is known as Euler’s Theorem. It is quite powerful as you will see.
Hint: Use LaGrange’s Theorem.
Solution: Since a is relatively prime to n,
a!U(n)
. Thus,
ao(U(n)) =1!a
"
(n)=1
.
