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$|G|=112$, prove that $G$ is solvable.

I already proved that $G$ is not simple so I know that it has normal subgroup but I don't know from what order. Another solution that I tried was looking at the sylow subgroups. In that case, if 2-sylow subgroup is normal or 7-sylow subgroup is normal then the solution is obvious, but in case that there are 7 2-sylow subgroups and 8 7-sylow subgroups, I don't know how to continue.

user7080065
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  • Well, what possible orders of a normal subgroup did you conclude? – Tobias Kildetoft Sep 28 '17 at 08:30
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    Consider the comjugation action of $G$ on the $7$ Sylow $2$-subgroups. The image is divisible by $7$. If it is not contained in $S_7$ then $G$ has a normal subgroup of index $2$. Otherwise the action is not faithful, and the kernel is a $2$-subgroup of $G$. Either way this reduces to showing groups of order $2^k\cdot 7$ are solvable for $k \le 3$. For $k\le 2$ this is easy. For $k=3$ see for example https://math.stackexchange.com/questions/353552 – Derek Holt Sep 28 '17 at 08:49

2 Answers2

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The number $112=2^4\cdot 7$ has prime factors $2$ and $7$. By Burnside's $p^aq^b$-theorem, all groups of such order, having only two different prime divisors, are solvable. For order $p^aq$ the proof is not difficult, see here.

In case you do not want to use Burnside's result, we already know from the other question,

Proving that a group of order $112$ is not simple

about the Sylow subgroups. For example, one possibility is that the group has a normal Sylow $7$-subgroup $N$ of order $7$. Hence the quotient has order $2^4$ and is nilpotent (since it is a $p$-group), hence solvable. Now, if $N$ and $G/N$ are solvable, so is $G$, and we are done.

Dietrich Burde
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  • I am not allowed to use Burnside's theorem and the solution for the sylow subgroups is quite a mess and that's exactly what I'm missing – user7080065 Sep 28 '17 at 09:03
  • But you can use the arguments from the proof for groups of order $p^aq$, which are not difficult (see the link). – Dietrich Burde Sep 28 '17 at 09:31
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Another approach is to show that all the composition factors of $G$ are of prime order. $A_5$ is the only non-abelian simple group of order $<100$: $G$ is non abelian simple group of order $<100$ then $G\cong A_5$

Since $G$ is not simple as you've claimed, the composition factors of $G$ have order not greater than $56$. Conclude that they are all abelian, and thus cyclic of prime order.

Well... the upper bound $100$ is more than enough. You only need to show that there is no non abelian simple group of order not greater than $56$.