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Show that if a subgroup $H$ of $A_4$ contains at least $7$ elements, then $H=A_4$.

I can show picking $7$ elements they generate $A_4$, but I don't know how to show it works in every case.

$$A_4=\{1, (123), (132) , (124) , (142) ,(134) , (143) , (234), (243) , (12)(34) , (13)(24) , (14)(23)\}.$$

  • 5
    Do you know that the order of a subgroup divides the order of the group? – 2017-02-23

2 Answers 2

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You know by Lagrange's theorem, that the order of a subgroup divides the order of the group.

So in your case,

$$\vert H\vert \mid \vert A_4\vert=\frac{4!}2=12.$$

So if $\vert H\vert\geqslant 7$, then $\vert H\vert=12$, so $H=A_4$.

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HINT:: The order of $A_4$ is $12$, while the order of $\langle H \rangle$ is at least 7. Assume $\langle H \rangle$ is a proper subgroup of $A_4$ and then use Lagrange's Theorem to obtain a contradiction.