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I was looking at this thread and it doesn't quite answer my question.

Classify all groups of order 182

Also, I am using Lang's Algebra. So far we have only covered the first 7 chapters.

Here is what I have.

Let $G$ be a group of order 182. Since $182=2\cdot7\cdot13,$ sylow's theorems guarantee the existence of p-sylow subgroups denoted by $H_2,H_7,H_{13}$ respectively. Again, by Sylow, we have that the number of sylow-7 subgroups, denoted by $n_7$, is 1 since $n_1\equiv1\mod7$ and $n_7\mid26$. Since $n_7=1$ and all p-sylow subgroups are conjugate, we get that $H_7$ is unique and thus for any $g\in G$ we have $gH_7g^{-1}=H_7$. Therefore, $H_7\lhd G$. Thus, $G/H_7$ is a group or order 26 and is isomorphic to a subgroup $A<G$. Since $A$ has order $26=2\cdot13$, it has order equal to the product of two distinct primes, and is therefore solvable. Since $G/A$ has order 7, it is cyclic, which means it is abelian, and thus it is solvable. Therefore, since $A$ and $G/A$ are solvable, $G$ is solvable.

My issue is, I just found out that if $H\lhd G$, then $G/H\cong K$ for some $K<G$ is not always true. That statement is true for abelian groups, but we don't have the assumption that $G$ is abelian. Is there a way to salvage this, or am I going about it in the wrong way?

JohnC
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    All groups of order $182$ are soluble. – Angina Seng Oct 12 '17 at 21:19
  • You don't need $A$ to be isomorphic to a subgroup of $G$. You have $H_7$, and normal cyclic (order $7$) subgroup of $G$ is solvable, and by your argument, $G/H_7$ is solvable, so $G$ is solvable. – James Oct 12 '17 at 21:21
  • For clarification, you are saying that even though $G/H_7$ is not isomorphic to a subgroup of $G$, it is still solvable because of its order. Thus, look at $G/(G/H_7)$ which is a group of order 7 so it is solvable. Therefore, $G$ is solvable? Is this right? How do you know that $G/H_7\lhd G$ if it may not even be a subgroup of $G$ – JohnC Oct 12 '17 at 21:28
  • @JohnC No. You've argued that $H_7$ is normal in $G$, and that $G/H_7$ is solvable. But $H_7$ itself is solvable (it is cyclic!). So you have that $G$ is solvable-by-solvable, hence, itself solvable. – James Oct 12 '17 at 21:41
  • Got it. The theorem I'm looking at doesn't mention that $G/H_7$ has to be a subgroup, but I believe I was thinking of the actual definition involving abelian towers and kept trying to put the quotient in the tower for whatever reason. – JohnC Oct 12 '17 at 21:53
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    The title of this question says the opposite of what you mean, so you should correct it. – Derek Holt Oct 13 '17 at 07:27

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So, in short, as $P\in Syl_7(G)$ and we have $P\unlhd G$ by divisibility condition on the no. of Sylow $7$-subgroups of $G$, we obtain our quotient group $G/P$, of order $2\times 13$. By Burnside, the quotient is solvable. Both $P$ and $G/P$ are solvable imply $G$ is solvable.