Math 546
Problem Set #11
In these and future problems when we say that G is a group and write xy for two
elements x, y of G we mean to imply that the operation of G is being applied. So if the
operation of G is * we might write xy instead of x * y . Also, if the operation is
commutative we might write x+y instead of x * y . In most cases we will not even use
a special symbol such as * to denote the group operation. In some special cases we
may still use notation such as x * y if we want to underscore the group operation.
1. Suppose that G is a group and x and y are elements of G. Suppose that n and m are
relatively prime integers and that x n = y n and xm = y m . Show that x = y.
Hint: x = x 1 . Now what significant theorem have we discussed that deals with
relatively prime integers?
Solution:
Since n and m are relatively prime, there exist integers a and b such that
an + bm = 1 . Hence,
x = x 1 = x an + bm = x an x bm = ( x n ) ( x m ) = ( y n ) (y m ) = y an + bm = y1 = y
a
b
a
b
If x is any element of the group G, and A is any subset of G, then
xA = { xa : a ! A } (this is a definition). Similarly, if A and B are any two subsets
of G, then AB = {ab : a ! A, b ! B} .
A commutative group is often said to be Abelian.
2. Let A and B be subgroups of the Abelian group G.
Show that AB = {ab : a ! A, b ! B} is a subgroup of G.
Solution: Let e denote the identity element of G. Since e belongs to both A and B,
then e = ee ! AB .
Claim: AB is closed.
Suppose that x and y are any two elements of AB. Then x = a1b1, y = a2 b2 for some
a1 ,a2 ! A b1, b2 !B . But then xy = a1b1 a2 b2 = ( a1 a2 )(b1b2 ) ! AB (note that we used the
commutativity of G here.
Claim: If x is in AB, then so is x !1 .
Let x ! AB . Then x = ab for some a ! A, b !B .
So, x !1 = (ab)! 1 = b!1 a!1 = a!1b !1 " AB (We also used the commutativity of G here.)
3. Suppose that H is a subgroup of G and x !H ,!and!y "H .
Show that xy !H . [This is the converse of a problem from the last problem set.]
Solution: Suppose that xy !H . Then since H is a subgroup, then (xy)!1 "H .
Thus since x is in H and H is closed under the operation on G, (xy)!1 x "H . But
since (xy)!1 x = y !1 x !1 x = y !1 , this implies that y !1 "H . But since H is a subgroup
and thus contains the inverse of each of its elements, y = (y !1 )!1 must belong to H.
But this is a contradiction since we were given that y !H .
Note that a similar argument to that above shows just as well that yx !H .
By the way, the next problem was once on a Putnam exam.
4. Prove: No group is the union of two proper subgroups.
Hint: Suppose that G = A ! B where G is a group and A and B are proper
subgroups of G. Then A cannot be a subset of B (why?) and similarly B cannot be
a subset of A. Hence we can choose a ! A " B …
Solution: Suppose that G = A ! B and that A and B are subgroups of G but that
neither is equal to G.
Solution: Suppose that G = A ! B and that A and B are subgroups of G but that
neither is equal to G. Then it must be that A ! B " # and B ! A " # . So now
choose a !A " B and b !B " A . Then since a !A and b !A , we get that
ab !A . Similarly, b !B and b "A and so ab !B .