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Show that if a finite group, $G$, contains a proper subgroup of index $2$ in $G$, then $G$ is not simple.

Proof

Let $H$ be a proper subgroup of index $2$ in $G$. We know that $H$ is normal because it is of index $2$. Now, assume that $H = \{e\}$. Thus, $G/H \simeq G$. However, $G/H$ is of order $2$ because $H$ is of index $2$. Since $H$ is a proper subgroup, the order of $H$ must be less than the order of $G$. This means that the order of $G$ is greater than $G/H$. Thus, $G$ is not isomorphic to $G/H$ and $H \neq \{e\}$ and $G$ is not simple.

Now, I have a problem with the statement "Since $H$ is a proper subgroup, the order of $H$ must be less than the order of $G$. This means that the order of $G$ is greater than $G/H$." If $|G|=2$, then the proof seemingly falls apart. How can I rectify this problem?

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    $\begingroup$ Your objection is valid. The cyclic group of order $2$ is simple, yet it has a proper subgroup of index $2$. I expect they meant to exclude that case. $\endgroup$ – lulu Jul 11 '20 at 16:52
  • $\begingroup$ @lulu So the statement I attempted to prove is invalid? $\endgroup$ – N. Bar Jul 11 '20 at 17:01
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    $\begingroup$ Yes, but only in that one exceptional case. Your argument still holds if $|G|>2$ or if you exclude the case in which $H$ is trivial. $\endgroup$ – lulu Jul 11 '20 at 17:08
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    $\begingroup$ @N.Bar And, to stress, your work is correct. You drilled down to the one exceptional case and you correctly handled the other cases. $\endgroup$ – lulu Jul 11 '20 at 17:14
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    $\begingroup$ @N.Bar Ah, sorry. Due to your direct quote after the proof, I assumed you were referring to somebody else's proof. In that case, no wonder you tried to elaborate finer and finer details because after all you simply ran against the one exceptional case where the given claim fails to be true. $\endgroup$ – Hagen von Eitzen Jul 11 '20 at 17:18
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As discussed in the comments, this works if and only if $|G|>2$. However, there are some authors who use the term "proper" to mean that $H\neq G$ and $H\neq \{e\}$. With this definition the statement in the problem is correct. You should check whether this is the definition being used in whatever source you found the problem.

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  • $\begingroup$ Thank you. My book defines proper as $H \neq G$. But this is definitely something that should be taken into account. $\endgroup$ – N. Bar Jul 11 '20 at 18:16

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