Subgroup of Abelian group

**tttcomrader** · August 28th 2007, 09:11 AM

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 $a,b \in G$ such that $<a> = 2, <b> = 2$ , so we have $a^{2}=e, b^{2}=e$ .

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

Am I starting this right?

**ThePerfectHacker** · August 28th 2007, 02:07 PM

Originally Posted by tttcomrader

Am I starting this right?

Try this.

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

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