# Subgroup of Abelian group

• Aug 28th 2007, 09:11 AM
Subgroup of Abelian group
Prove that an Abelian group with two elements of order 2 must have a subgroup of order 4.

My proof:

Suppose that G is an Abelian group, and let $\displaystyle a,b \in G$ such that $\displaystyle <a> = 2, <b> = 2$, so we have $\displaystyle a^{2}=e, b^{2}=e$.

From a theroem, we know that <a> and <b> are subgroups of G.

Am I starting this right?
• Aug 28th 2007, 02:07 PM
ThePerfectHacker
Quote:

We know that $\displaystyle G$ has two elements $\displaystyle a\mbox{ and }b$ which has orders two. This means $\displaystyle a^2 = b^2 = 1$. Now consider the elements: $\displaystyle 1,a,b,ab$. Are these elements distinct? Well, $\displaystyle a\not =1 \mbox{ and }b\not =1$ (why not?). And $\displaystyle a\not = ab \mbox{ and }b\not = ab$ (why not?). And finally can $\displaystyle ab=1$? It turns out that no (why not?). If you can show that all these elements are distinct then form the subset $\displaystyle H= \{ 1 , a , b , ab\}$. Show that this set is a group. And hence a subgroup of order 4.