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 \$\displaystyle \mathbb{Z}_2\$ but in the spirit of your question...no. If \$\displaystyle G\$ is a finite group and \$\displaystyle |G|=21\$ where \$\displaystyle m>1\$ then a simple application of Cauchy's theorem says that \$\displaystyle G\$ must have an element of order \$\displaystyle 2\$. More generally, Sylow's first theorem implies that if \$\displaystyle G\$ is a finite group and \$\displaystyle p^m\mid |G|\$ with \$\displaystyle p\$ prime then \$\displaystyle G\$ must contain a subgroup of order \$\displaystyle p\$. From this it's easy to see that the only groups you mention have order \$\displaystyle 1\$ or \$\displaystyle p\$ where \$\displaystyle 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 \$\displaystyle \mathbb{Z}_2\$ but in the spirit of your question...no. If \$\displaystyle G\$ is a finite group and \$\displaystyle |G|=2m\$ where \$\displaystyle m>0\$ then a simple application of Cauchy's theorem says that \$\displaystyle G\$ must have an element of order \$\displaystyle 2\$. More generally, Sylow's first theorem implies that if \$\displaystyle G\$ is a finite group and \$\displaystyle p^m\mid |G|\$ with \$\displaystyle p\$ prime then \$\displaystyle G\$ must contain a subgroup of order \$\displaystyle p\$. From this it's easy to see that the only groups you mention have order \$\displaystyle 1\$ or \$\displaystyle p\$ where \$\displaystyle p\$ is prime.

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

-Dan