Counterexample of the reciprocal of Langrage's theorem.

Asked Jan 11 '20 at 14:11

Active Jan 11 '20 at 14:36

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Lagrange's theorem states that if $G$ is a finite group, the order of every subgroup $H<G$ must verify that $|H| \mid |G|$ (the order of $H$ divides the order of $G$)

The reciprocal result is not true, but the only example I could find is the group of even permutations $A_4,$ which has order $12$ and does not have any subgroup of order $6.$

Is there any other 'easy' counterexample of this?

edited Jan 11 '20 at 14:36

Shaun

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asked Jan 11 '20 at 14:11

Somerandommathematician

I think this the easiest example. – math Jan 11 '20 at 14:21
I know, actually i think that $A_4$ is the smallest group where you can find a counterexample, but I'm interested in any other one. – Somerandommathematician Jan 11 '20 at 14:22
check this one,https://math.stackexchange.com/questions/103266/counterexamples-to-converse-to-lagranges-theorem-different-from-a-4 – math Jan 11 '20 at 14:25
Oh wow, that was helpfull. Thank you. – Somerandommathematician Jan 11 '20 at 14:27
yes indeed, is what @math commented before. – Somerandommathematician Jan 11 '20 at 14:47

Counterexample of the reciprocal of Langrage's theorem.

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