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I have a question I was hoping for help on:

Prove or disprove every nonabelian group of order divisible by 6 contains a subgroup of order 6

I would guess that this statement is true based on a few examples I've done, but I have no idea why (or if it really is true). If I had a guess there might be some sort of isomorphism that I can relate non abelian groups of order 6 to. Would someone be able to help me? If you wouldn't mind, if you can help me, provide a simplistic explanation so that I may understand better. Thanks in advance for your help, I really appreciate it!

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The statement is not true.

You can verify that $A_4$ is a counterexample.

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  • $\begingroup$ could {e,(12),(13),(23),(123),(132)} be considered a subgroup of order 6 in $S_4$? $\endgroup$ – Billy Thorton Nov 28 '14 at 19:50
  • $\begingroup$ Yes, I am not saying thet $S_4$ is the counterexample, I am saying that a subgoup of $S_4$ is. $\endgroup$ – Pavel Čoupek Nov 28 '14 at 20:02
  • $\begingroup$ I apologize, I'm not exactly sure what you mean. Do you mean the subgroup $S_4$ as a part of another, larger symmetric group? Say like $S_5$? $\endgroup$ – Billy Thorton Nov 29 '14 at 3:33
  • $\begingroup$ @BillyThorton: Exactly the opposite. Look at $A_4$ and show that this is the counterexample. $\endgroup$ – Pavel Čoupek Nov 29 '14 at 7:32
  • $\begingroup$ So, just for my understanding, the set I would examine in $A_4$ and the fact that it is a subgroup of $S_4$, then the counterexample would follow? $\endgroup$ – Billy Thorton Nov 29 '14 at 15:47

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