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Let $G$ be this group and $H$ be any subgroup of index 4.

$G$ acts on the set of left cosets of $H$ in $G$, which is a homomorphism $\varphi: G\to Aut(G/H) = S_4$. It is easy to see that $\ker \varphi\subset H$. We obtain that $$|G/\ker\varphi| = [G:\ker\varphi] = [G:H][H:\ker\varphi] = 4[H:\ker\varphi].$$ But $G/\ker\varphi \hookrightarrow S_4$, thus $|G/\ker\varphi|$ divides $4! = 24$. Therefore $G/\ker \varphi$ can have order $4,8, 12, 24$.

$G/\ker \varphi$ cannot have order $8$, since by Sylow first theorem we have a normal subgroup of order 4, which is index 2. This subgroup lifted to $G$ will be a subgroup of index 2. $G/\ker \varphi$ cannot have order 12, since the Sylow subgroup of 2 is index 3 and can be lifted to a subgroup of index 3 in $G$. $G/\ker \varphi$ cannot have order 24 for similar reason, or one can argument $S_4$ has $A_4$ sitting in it.

Therefore $G/\ker \varphi$ has order 4, meaning $H = \ker \varphi\lhd G$. Please correct me if I am wrong.

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  • $\begingroup$ Your argument is correct. $\endgroup$
    – mesel
    Jun 14 '14 at 17:51
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This, of course, never happens: suppose $\;H\le G\;,\;\;[G:H]=4\;$ . As you proved, we have that $\;H\lhd G\;$, but then no matter what of the two groups of order four is $\;G/H\;$ , it will have a subgroup $\;K/H\;$ of order two, with $\;H\le K\le G\;$ , and by the correspondence theorem we then have that

$$[G/H:K:H]=2\iff [G:K]=2$$

and we thus have a subgroup of index two...

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  • $\begingroup$ That's true. So if a group has no subgroup of index 2 or 3, then it does not have any subgroup of index 4. $\endgroup$
    – mez
    Jun 15 '14 at 11:54
  • $\begingroup$ Indeed so, @mezhang. This, in fact, also follows directly from Cauchy's Theorem: if we have a subgroup of order four then two divides the group's order and thus there's an element of order two in the group... $\endgroup$
    – DonAntonio
    Jun 15 '14 at 11:58
  • $\begingroup$ Yea, but what you said only applies to finite groups, this works for any group. $\endgroup$
    – mez
    Jun 15 '14 at 12:58
  • $\begingroup$ Indeed so, @mezhang: Cauchy's theorem is for finite groups. $\endgroup$
    – DonAntonio
    Jun 15 '14 at 14:20
  • $\begingroup$ Wait, regarding your previous comment, I don't see how Cauchy's Theorem deduces the statement "If a group has no subgroup of index 2 or 3, then it does not have any subgroup of index 4". What does an element of order 2 lead us to conclude? $\endgroup$
    – mez
    Jun 15 '14 at 18:25

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