Subgroup of order 2 of a group of order 56

Question

I was given the following question: Does every group of order 56 contain a subgroup of order 2. I know that the Sylow theorems guarantee the existence of an order 8 subgroup. Is there a general technique to prove the existence or the non-existence of subgroups of lower prime powers? Thank you for your help.

Answer 1

As you say, the first Sylow theorem tells you that there is a subgroup of order $8$. Take a non-identity element $g$ of that subgroup. It has either order $2, 4$ or $8$, by Lagrange's theorem. If $g$ has order $2$, we're done. If $g$ has order $4$, then $g^2$ has order $2$, and we're done. If $g$ has order $8$, then $g^4$ has order $2$, and we're done.

Answer 2

Hint : It is well known that if a prime $p$ divides the order of a group $G$ , an element of order $p$ in $G$ exists. Take it from here.

Answer 3

A well-known fact from group theory is that a group of order $p^n$ has subgroups of order $p^m$ for all $0\leq m\leq n$.

Proof. By induction on $n$. A non trivial subgroup has non trivial center. Pick $z$ in the center, apply induction to $G/\langle z\rangle$, and lift your subgroups to $G$ (that is take their inverse images by the canonical projection).

Applying this result to Sylow subgroups gives you the following theorem.

If a finite group $G$ has order divisible by $p^n$, it has subgroups of order $p^m,$ for all $ 0\leq m\leq n$.

Answer 4

Let $G$ be a group of order $56$. If $g\in G$ has even order $2n$, then the order of $g^n$ is $2$, and you are done. What if every element has odd order? Since the order if each element must divide $56$, that would mean that every $g\in G\setminus\{e\}$ must have order $7$. But the intersection of any two distinct subgroups of order $7$ is $\{e\}$. So, since $G$ coud be written has the union of subgroups of order $7$ and since any two such subgroups, if they are distinct, only share $e$ as a common element, $56$ could be written has $1+6k$, for some $k\in\mathbb N$. However, no such $k$ exists.