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Can anyone check my work for two questions I have attempted as I'm not sure it is correct:

1)

I want to show every group of order $12$ is solvable. Here is what I attempted.

It is easy enough to show that either the Sylow $2$ subgroup is unique or the Sylow $3$ subgroup is unique.

Suppose the Sylow $2$ subgroup is unique: Let $H$ denote this subgroup. We note that $H$ is normal since it is unique. Then we get the following series $1 \trianglelefteq H \trianglelefteq G$. Then $G/K$ has size $3$ and is solvable since all $p$ groups are solvable. Likewise $H/1 \cong H$ has order $4$ and is again solvable. So $G/H$ and $H$ are both solvable groups so we deduce $G$ is solvable.

If the Sylow $3$ subgroup is unique we can apply the same argument where the subgroup has size $3$ and the quotient has size $4$ and in this case $G$ is also solvable.

I know I am skipping a few details but I want to check that my argument makes sense. Thanks!

2)

We claim all groups of order $28$ are solvable.

Consider Sylow $7$ subgroups. Let $n$ be the number of these subgroups then $n\equiv 1 \mod 7$ and $n|4$ so $n=1$ hence there is a unique (and hence normal) Sylow $7$ subgroup. Let $K$ denote this subgroup.

Again we consider the normal series $1 \trianglelefteq K \trianglelefteq G$ we have $K$ is solvable (order $7$) and $G/K$ is solvable (order $2^2$). So we get that $G$ is solvable???

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Yes, you are correct. The general fact is as follows:

Let $G$ be a group and $N$ its normal subgroup. Then $G$ is solvable if and only if both $N$ and $G/N$ are solvable.

Now $p$-groups are solvable. That's because every $p$-group is either abelian or has a proper, nontrivial normal subgroup (the center). So by induction (on the order of $G$) it has to be solvable.

So the only thing left to prove in your case is that in 1) you have to show that a group of order $12$ has a normal Sylow subgroup. But if you say you know how to do it then entire reasoning is correct.

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There are probably many ways to show the result. I think your proof is essentially correct, some details have to be worked out. Of course you could show more generally that all finite groups of order $p^2q$ are solvable, for primes $p,q$.

Here is another possible proof for $|G|=12$. The number $s_3$ of Sylow $3$ subgroups is $1$ or $4$. Suppose first that $n_3=4$: consider the conjugation action of $G$ on the set of Sylow's $3$- subgroups. It yields a homomorphism $$ \phi\colon G\rightarrow S_4, $$ As the action is transitive, $|im(\phi)|\ge 4$. If $|im(\phi)|= 4$ then $\ker(\phi)$ would be a normal subgroup of order $3$ which contradicts the assumption that $n_3 = 4$. Thus, $|im(\phi)|= 12$, so $\phi$ is injective. The only subgroup of $S_4$ of order $12$ is $A_4$. Therefore, $G\cong A_4$, which is solvable.

For $s_3=1$ we either have $s_2=1$, in which case $G$ is abelian, or $n_2=4$, which is either $G\cong S_3\times C_2$ or another solvable group of order $12$.

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