Expert: Paul Klarreich - 2/23/2007

Question
Hi,

i study discrete math

Would you please help me with the following 2 questions

1-Prove that every infinite group contains infinitely many subgroups.

2- If a1, a2 . . . , an are (not necessarily distinct) elements in a group G, prove that
(a1a2.....an)^(-1)=(an)^n.....(a2)^-1(a1)^-1

Answer
Questioner:   Linda
Category:  Advanced Math
 
Subject:  Groups
Question:  Hi,

i study discrete math

Would you please help me with the following 2 questions

1-Prove that every infinite group contains infinitely many subgroups.

2- If a1, a2 . . . , an are (not necessarily distinct) elements in a group G, prove that
(a1a2.....an)^(-1)=(an)^n.....(a2)^-1(a1)^-1

>> Typo? You meant (an)^-1, right?

.....................................
Hi, Linda,

I don't have a proof right now for part 1, and I'll think about it some more, and possibly send it on to my 'supervisor' expert.

Part 2 can be proved using mathematical induction.  If you don't have a lot of practice with that, the steps are:

A. Prove the theorem for a basic case, like n=1 or n=2.
B. Assume the theorem true for n=k.
C. Use that assumption to prove it for n=k+1.

Base case: Prove that  (ab)' = b'a'
[I'll just use a,b instead of a1,a2 here.   And I will write ' instead of -1 for the inverse, to save typing and make things easier to read.]

b'a'(ab) = b'(a'a)b  [associative]
= b'(e)b  [definition of inverse]
= b'b  [definition of unity]
= e [inverse]

Now assume the theorem for  n = k, which means:

(a1 a2 .. ak)' = ak' ... a2' a1'  [Assumption.]

Next prove the theorem for n = k+1

(a1 a2 .. ak ak+1)' = ak+1' ak' ... a2' a1'   << to be proved.

(a1 a2 .. ak ak+1)' = ((a1 a2 .. ak) ak+1)'  [Assoc.]
= ak+1' (a1 a2 .. ak)' [Base case]
= ak+1' ak' ... a2' a1'  [By assumption]