Group of order $224$

Solution 1:

Extended hints:

Let $P$ by a Sylow $7$-subgroup. There are two alternatives for $n_7$, the number of distinct conjugates of $P$ (=all the Sylow $7$-subgroups). Find them.

• If $n_7>1$, then what can you say about the normalizer $N_G(P)$?
• If $n_7=1$, then can you use the fact that $p$-groups always have subgroups of all orders allowed by Lagrange's theorem? Recall that a product of two subgroups is a subgroup, if one of them is normal.

Solution 2:

Let $P \in Syl_7(G)$, then, since $|Syl_7(G)|=[G:N_G(P)]\equiv 1 \text { mod } 7$, there are two cases

(a) $[G:N_G(P)]=1$
(b) $[G:N_G(P)]=8$.

In case (a), $P \trianglelefteq G$. Choose a subgroup $H/P$ of $G/P$, such that $|H/P|=4$. Such a group exists, since $G/P$ has order 32 and has a subgroup of order 4 (in fact for any power of $2$ up to $32$). Then $|H|=28.$

In case (b), observe that $|N_G(P)|=28$.