# Group of order $224$

Any one can give a hint to prove this?

Every group of order $224$ have a subgroup of order $28$.

• Did you have any luck working towards an answer yourself? Maybe you know something that could serve as a hint to which techniques are called for? Commented Aug 13, 2014 at 14:04
• I know that there is a group of order $7$ and a group of order $32$, using the sylow theorems Commented Aug 13, 2014 at 14:10

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.
• What prevents $|N_G(P)|=14$ ? Commented Aug 13, 2014 at 14:11
• @Rene: $n_7=[G:N_G(P)]$. If $N_G(P)=14$, then $n_7$ would be $16$. Commented Aug 13, 2014 at 14:12
• Mind you, the second bullet is a bit overkill. The required other subgroup can be constructed either as a subgroup of the center of a Sylow subgroup $Q$, or by using the correspondence on $Q/Z(Q)$. Commented Aug 13, 2014 at 14:16
• sorry I dont understand the comment $Q$ is the $2$-sylow subgroup ? What prevents its center from having order $2$ ? Commented Aug 13, 2014 at 14:18
• So, in the first case $N_G(P)$ has order $28$, and in the second case i am not sure, you say to take a group order $4$ and take the product with $P$? Commented Aug 13, 2014 at 14:39

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 $|N_G(P)|=28$.

• @Babak, sure go ahead! Commented Aug 13, 2014 at 20:22
• Dear Nicky, it's posted. :) Commented Aug 17, 2014 at 7:50

Let $G$ be a group of order $224=2^5 \cdot 7$.

Let $n_2$ be the number of Sylow $2-$subgroups in $G$ and $n_7$ be the number of Sylow $7-$subgroups in $G$.

$n_2=1 \text{ or } 7$

Using the third Sylow theorem, $n_2=1 \text{ or } 7$, $n_7=1 \text{ or } 8$.

Suppose that $n_2=7$ and let $P_2$ be a Sylow $2-$subgroup. Therefore, $7=n_2=[G:N(P_2)]=\frac{|G|}{N(P_2)}=\frac{224}{N(P_2)}$, where $N(P_2)$ is the normalizer of $P_2$ in $G$. So, $|N(P_2)|=32$.

Now suppose that $n_2=1$. Let $P$ be the unique Sylow $2-$subgroup of order $2$, and let $Q$ be a subgroup of order $7$. Since $P$ is normal in $G$ and $P \cap Q=\{e\}$, $PQ$ is a subgroup of $G$ of order $14$.

Suppose that $n_7=8$ and let $P_7$ be a Sylow $7-$subgroup. Therefore, $8=n_7=[G:N(P_7)]=\frac{|G|}{N(P_7)}=\frac{224}{N(P_7)}$, where $N(P_7)$ is the normalizer of $P_7$ in $G$. So, $|N(P_7)|=28$, so $N(P_7)$ is the desired subgroup.

Now suppose that $n_7=1$. Let $P$ be the unique Sylow $7-$subgroup of order $7$, and let $Q$ be a subgroup of order $2$. Since $P$ is normal in $G$ and $P \cap Q=\{e\}$, $PQ$ is a subgroup of $G$ of order $14$.

Therefore, every group of order $224$ has a subgroup of order $28$.