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If a group of order $168$ has more than one sylow $7$ subgroups then is it simple?

I am relatively new to solving problems like this. I tried solving this problem as follows:

Let us assume that $G$ is not simple. My strategy was to assume that $\exists H\unlhd G$ such that $H\neq {e},G.$

$$168=7.3.2^3$$

If $7||H|,$ then $|H|=7,14,21,28,42,56,84,168.$ If $|H|\neq 56,168$ then $H$ has a unique Sylow 7 subgroup which is also a characteristic subgroup of $H$ which means that $G$ has a normal and hence, a unique Sylow-7 subgroup, a contradiction. Also, $H\neq 168$ as $H\neq G$ so, $|H|=56$ is the only case left, when $7||H|.$ We note that $[G:H]=3\implies $ $G/H$ is cyclic.

If the number of Sylow-7 subgroups in $H$ is 8 then, it must have a unique Sylow 2 subgroup and this implies the number of Sylow $2$ subgroups in $G$ will be $1$ as well.

If somehow, I can show that $|H|$ cannot be 56 as well, then we need to consider more cases like when $3||H|$ and $2||H|.$ But then the solution becomes extrememly lengthy. And I don't know if it will bear any fruit by continuing it this way. I need some help to proceed further.

Thomas Finley
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This is not true.

Consider $C_3\times [(C_2)^3 \rtimes C_7]$ .

ancient mathematician
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    A more exotic counterexample would be the extension of the additive group of $\mathbb{F}_8$ by the non-abelian group of order $21$ acting as semi-linear transformations. – ancient mathematician May 07 '25 at 14:46
  • What will be the homomorphism $\phi$ in $(C_2)^3\rtimes_{\phi} C_7$? – Thomas Finley May 09 '25 at 12:10
  • Any non-trivial one. The companion matrix $C(X^3+X+1)$ in $GL(3,2)$. – ancient mathematician May 09 '25 at 12:39
  • I am sorry, but I don't know what is a companion matrix. What I thought was that we can map the generator of $C_7$ to an element of order $7$ in $Aut((C_2)^3)$ and such an element exists by Cauchy's Theorem. But how to show that that mapping is precisely a homomorphism? – Thomas Finley May 09 '25 at 12:42
  • Thomas: If $C_7=\langle x\rangle$, and $y$ is of order $7$ in $GL(3,2)$ (or $Aut(C_2^3)$ if you prefer) then surely it is not too difficult to see that $x^m\mapsto y^m$ is a well-defined 1--1 onto homomorphism. – ancient mathematician May 09 '25 at 14:41
  • Oh, I get it, as $\phi(x^mx^n)=\phi(x^{m+n})=y^{m+n}=y^my^n=\phi(x^m)\phi(x^n),$ (where $x^m,x^n$ are arbitrary elements of $$), right? – Thomas Finley May 09 '25 at 15:37
  • That does the homomorphism bit, but it's important to prove that the map is well-defined and also 1--1 – ancient mathematician May 09 '25 at 15:40
  • I get that we need to prove the well-definedness. But why $1-1$? Whether $\phi$ is 1-1 or not seems to be not of interest in this problem, isn't it? But how to show the well-definedness? I think well-definedness is kinda obvious as we are mapping a genrator of a group to a unique element of another group. Isn't that sufficient? – Thomas Finley May 09 '25 at 15:47
  • If it were not 1--1 we would have the trivial action which is not what we want. "Kinda obvious" just doesn't do. We have defined the image of $p=x^n$ in terms of $n$ so we need to check that if $p=x^m=x^n$ we get the same image for $p$ both ways. – ancient mathematician May 09 '25 at 15:53
  • If it were not 1--1 then how would we have the trivial action? The trivial action means $\phi(x)=e$, isn't it? For in this case, $\phi$ maps every element of $$ to the identity of the other group. But if we have said that $\phi(x)=y$ where $o(y)=7$ then how can $\phi$ be trivial? It's non trivial in this case. Am I missing something? – Thomas Finley May 09 '25 at 16:17
  • Did I make any sense in the above comment? – Thomas Finley May 10 '25 at 05:29
  • You are giving the right reasons, I am trying to tell you why you need to give them! – ancient mathematician May 10 '25 at 07:04
  • Thanks a lot! I do get it now. +1 from me. – Thomas Finley May 10 '25 at 15:40