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Practice Question Prove that a group with two elements of order 2 that commute must have a subgroup of order 4?

I went with the approach that a group with order 2, can generally not exist under group axioms unless there is a subgroup with order 4. Friends and I were almost positive this can be proven by contradiction. Suggestions? Solutions? Different approaches?

Aspiring Mathematician
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  • The group cannot have order 2, because in addition to the two elements that commute it must also have an identity element. That's at least three elements. Be careful. – TBrendle Sep 28 '13 at 22:05

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HINT: There’s only one reasonable thing to try first: let $x$ and $y$ be the two elements of order $2$, and consider $\{1_G,x,y,xy\}$, where $1_G$ is the identity element of the original group $G$.

Brian M. Scott
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  • We should note that this will be a direct proof, not a proof by contradiction. – TBrendle Sep 28 '13 at 21:58
  • @TBrendle: Actually, I was going to let the OP figure that out from the hint. – Brian M. Scott Sep 28 '13 at 22:00
  • That's where I was headed, I simply figured my strategy can be, in a sense a contradiction. I attempted to word my proof by contradiction specifically, to generally satisfy meticulous critics and become a better mathematician. Thank you for your help. More questions are on the way. – Aspiring Mathematician Sep 28 '13 at 22:15
  • @PythagorasWannaBe: In this case it really is easier just to construct the group of order $4$; recasting the argument as a proof by contradiction just muddies the water a bit. // You’re welcome. – Brian M. Scott Sep 28 '13 at 22:17