# Subgroups of a group of even order

• Apr 3rd 2011, 02:52 PM
topsquark
Subgroups of a group of even order
My question is this: Can we construct a group of even order that has no subgroups? I am tolerably certain that I have heard of this as a theorem, but I may have only heard it in reference to cyclic groups where it is obvious that a subgroup of order 2 exists.

Thanks!
-Dan
• Apr 3rd 2011, 04:21 PM
Drexel28
Quote:

Originally Posted by topsquark
My question is this: Can we construct a group of even order that has no subgroups? I am tolerably certain that I have heard of this as a theorem, but I may have only heard it in reference to cyclic groups where it is obvious that a subgroup of order 2 exists.

Thanks!
-Dan

I assume that you mean no non-trivial proper subgroups. The answer is yes $\mathbb{Z}_2$ but in the spirit of your question...no. If $G$ is a finite group and $|G|=21$ where $m>1$ then a simple application of Cauchy's theorem says that $G$ must have an element of order $2$. More generally, Sylow's first theorem implies that if $G$ is a finite group and $p^m\mid |G|$ with $p$ prime then $G$ must contain a subgroup of order $p$. From this it's easy to see that the only groups you mention have order $1$ or $p$ where $p$ is prime.
• Apr 3rd 2011, 04:41 PM
topsquark
Quote:

Originally Posted by Drexel28
I assume that you mean no non-trivial proper subgroups. The answer is yes $\mathbb{Z}_2$ but in the spirit of your question...no. If $G$ is a finite group and $|G|=2m$ where $m>0$ then a simple application of Cauchy's theorem says that $G$ must have an element of order $2$. More generally, Sylow's first theorem implies that if $G$ is a finite group and $p^m\mid |G|$ with $p$ prime then $G$ must contain a subgroup of order $p$. From this it's easy to see that the only groups you mention have order $1$ or $p$ where $p$ is prime.

Thanks. "Cauchy" isn't until the next chapter, but I can follow the logic easily enough. Thanks!

-Dan