Let $G$ be a group of order $$|G|=\prod^n_{i=1}p_i$$ where each $p_i$ is a prime number and $$n\geq2\qquad \text{and} \qquad k\neq j\implies p_k\neq p_j \qquad \text{and} \qquad p_1<p_2<\cdots<p_n$$ I want to prove that $G$ is non-simple. I have tried to use Sylow's Theorem, some direct product tools but nothing. Is there some other theorem or method that can help me to prove it?
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1I take it you meant to exclude the case where $n=1$? It is the case that all groups of square free order are solvable, See, e.g., this – lulu Sep 10 '23 at 13:29
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This group is a semi-direct product of cyclic groups (this might help) You could try with the dihedral group first (with $p>2$ prime)? – julio_es_sui_glace Sep 10 '23 at 13:32
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Every finite group with an order not divisible by $4$ is solvable , in particular not simple. – Peter Sep 10 '23 at 13:50
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@lulu thanks, I forgot the case n=1, I fix that – Joao Sager Sep 10 '23 at 14:00
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@user10354138 thanks, I will read it, it's looks like very intersting – Joao Sager Sep 10 '23 at 14:01
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@Peter why? I didn't find de proof of that – Joao Sager Sep 10 '23 at 14:02
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The condition $p_1<p_2<\cdots<p_n$ makes redundant $k\neq j\implies p_k\neq p_j$. – citadel Sep 10 '23 at 15:24
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@citadel Hi, I now that, but I wrote that when I was trying to solve it using the Sylow Theorems, because you use that for element count – Joao Sager Sep 11 '23 at 12:56
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This theorem is what you need:
Theorem: Let $G$ be a finite group, and let $p$ be the smallest prime divisor of $|G|$. If a Sylow $p$-subgroup of $G$ is cyclic, then $G$ has a normal $p$-complement.
It follows immediately from this theorem that a finite group of square-free order cannot be simple, since all of its Sylow subgroups are cyclic.
A normal $p$-complement is a normal subgroup $N\triangleleft G$ such that $p$ does not divide $|N|$ and $|G:N|$ is a power of $p$. For example, if a group $G$ has order $60=2^2\cdot 3\cdot 5$, then a normal $2$-complement of $G$ would have $15=3\cdot 5$ elements.
The theorem I cited above is not easy to prove. The only proof I know uses Burnside's normal $p$-complement theorem, which is usually proven using transfer theory or character theory.
semisimpleton
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