All groups of order $112$ have an element of order $14$

Question

In finite groups there exists a very important class as Frobenius groups. We know that there exists a Frobenius group as $2^3:7$ which has an elementary abelian $2$-group. By the Properties of Frobenius groups there is no Frobenius group of order $2^4\times 7$, Since $ 7 \nmid 16-1$. When I checked by GAP it seems that all groups of order $112= 2^4\times 7$ have an element of order $14$. Could you please help me about the proof of it?

Answer 1

Let $G$ be a finite group of order $|G| = 112 = 2^4 \cdot 7$. Let $R$ be a Sylow $7$-subgroup. It suffices to prove the following

Claim: $R$ centralizes a non-trivial $2$-subgroup of $G$.

(This is because a commuting product of elements of order $2$ and $7$ has order $14$.)

If $R$ is normal in $G$, the claim follows from the fact that $Aut(R) \cong C_6$. So we will assume that $R$ is not normal in $G$. By Sylow it follows that $|N_G(R)| = 14$. We can assume that $N_G(R)$ is not cyclic, in which case $N_G(R)$ is dihedral of order $14$.

Note that $G$ is solvable, by Burnside's $p^a q^b$-theorem or as in here for example.

Thus a minimal normal subgroup $N \neq 1$ of $G$ is elementary abelian, and a $2$-group. If $|N| \not\equiv 1 \mod{7}$, the action of $R$ on $N$ by conjugation has a nontrivial fixed point, and we are done.

The only case that remains is $|N| = 8$, so $N \cong C_2 \times C_2 \times C_2$.

The action of $R$ on $N$ is faithful, so a generator of $R$ maps to an element of order $7$ in $Aut(N) \cong GL(3,2)$. Now note that because $N_G(R)$ is dihedral, a generator of $R$ is conjugate to its inverse in $G$. But in $GL(3,2)$, an element of order $7$ is not conjugate to its inverse, which can be seen by a bit of linear algebra.

In any case we have a contradiction, which completes the proof.

Other facts you can prove similarly:

• Every group of order $112$ has a normal Sylow subgroup.
• Every group of order $112$ has an element of order $2$ in its center.