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As seen in this answer, a group of order 144 is not simple. Now, I understand the main part of the answer, i.e. where it is concluded that, upon deducing that $n_3 = 16$, it is forced that $n_2 = 1$, and hence the group is not simple.

My questions:

  1. Why does $n_3=4$ imply that $G$ maps nontrivially to $S_4$? And why does this imply that $G$ is not simple?

  2. In the part "so by Sylow's theorem it has at least 4 Sylow 3-subgroups, and hence has order at least 36, so $|G:N_G(T)|≤4$ and G cannot be simple.", what exactly is the Sylow Theorem being applied to, and how is it used to deduce (using the fact that $|G:N_G(T)|≤4$) that $G$ is not simple?

  3. Are there other techniques besides the counting argument and the non-trivial mapping argument of $G$ to $S_n$ to show that $G$ of a certain order is not simple (using the Sylow theorems, that is)?

Any help would be much appreciated.

Community
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chubbycantorset
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3 Answers3

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There are certainly techniques of using theory of Sylow to prove non-simplicity; I would say the very first chapter of Isaacs Finite group theory shows these techniques (for example, uniqueness of Sylow subgroup will certainly imply non-simplicity; but if there are more than one Sylow subgroups, what is the nature of intersection of Sylow subgroups? Isaacs explores this a little bit, and also gives exercises based on this.

On the other hand, you may tray to search for how groups of order $p^nq$ can be shown to be simple? It certainly need tools from Sylow intersection (if one if of course not using character theory).

p Groups
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  1. $G$ acts on the set of Sylow 3-groups by conjugation, so there is a homomorphim $G$ to $S_4$.

  2. Again $G$ acts on the set of cosets of $N_G(T)$ by translation, so there is a homomorphim $G$ to $S_4$.

Boris Novikov
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It is kind of a counting argument, but in the cases where you cannot conclude that the Sylow subgroups intersect trivially, it can be useful to consider the normalizer of the intersection of two.

For example, if $G$ has order $36$, then it could have $1$ or $4$ Sylow $3$-subgroups. If it has $4$ and they do not all intersect trivially, let $H$ and $K$ be so $H \cap K$ is non trivial. Let $N(H\cap K)$ denote the normalizer of $H$ and $K$. Then the index of $H \cap K$ in $N(H \cap K)$ is congruent modulo $3$ to the index of $H \cap K$ in $G$. In this particular case, this implies that $3$ is a divisor of $[N(H \cap K) : H \cap K]$.

From this you can conclude that $|N(H \cap K)|$ is $18$ or $36$, which implies that $G$ is not simple.

Wow, only just realized this question was fairly old. Guess it got bumped. Anyway, maybe somebody will still find this useful.

NotAwake
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