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