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To be clear: I'm not asking whether $H$ is normal in $G$ or not. It is well-known that if such $H$ exists, then $H$ is normal in $G$, which is out of my interest right now.

But can we say that such $H$ always exists? (for any finite group $G$?)

Springfield
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    Not every finite group $G$ has non-trivial normal subgroup, so your statement can't be true. – Zerox Jan 01 '22 at 08:21
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    as a generalization to @Dovahkin's answer, $A_n$ ($n\geq 4$) cannot have a subgroup of index 2. https://math.stackexchange.com/questions/1501876/show-a-n-has-no-subgroups-of-index-2 – cineel Jan 01 '22 at 10:57

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$A_{4}$ does not have a subgroup of order $6$.

To elaborate , $A_{4}=2^{2}\cdot 3$

Since it does not have a subgroup of order $6$ . We cannot say that it has a subgroup of index $2$ which is the smallest prime dividing the order of $A_{4}$.

Dovahkiin
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