# abstract algebra proof

• March 3rd 2010, 05:26 AM
marches
abstract algebra proof
Can anyone help with the following?
Prove that G cannot have a subgroup H with the order of H=n-1, where the order of G=n, and n is greater than 2.
Thanks!
• March 3rd 2010, 05:53 AM
tonio
Quote:

Originally Posted by marches
Can anyone help with the following?
Prove that G cannot have a subgroup H with the order of H=n-1, where the order of G=n, and n is greater than 2.
Thanks!

Lagrange's Theorem: when $n-1\mid n$ ?

Tonio
• March 3rd 2010, 06:34 AM
marches
Thanks, but can't use Lagrange's...not to mention, G and H are infinite--I think Lagrange applies to finite groups. Any other suggestions?
• March 3rd 2010, 07:17 AM
tonio
Quote:

Originally Posted by marches
Thanks, but can't use Lagrange's...not to mention, G and H are infinite--I think Lagrange applies to finite groups. Any other suggestions?

You gave both G, H FINITE groups! You said the order of H = n-1 and the order of G = n...of course Lagrange applies only in the finite case: the case you gave.

Tonio
• March 3rd 2010, 07:27 AM
marches
oops--Right,right, I was thinking the n>2 thing was implying infinite order. But still, we were told not to use Lagrange's. Any other ideas?
• March 3rd 2010, 08:23 AM
Drexel28
Quote:

Originally Posted by marches
oops--Right,right, I was thinking the n>2 thing was implying infinite order. But still, we were told not to use Lagrange's. Any other ideas?

We have that $G-H=\{m\}$. Now, clearly $m\ne e$. Now, since $n>2$ we see that $|H|>1$ and so there exists some $h\in H-\{e\}$. Clearly $mh\in G$ but $mh\notin H$ since if it were we'd have that $mh=h',\text{ }h'\in H$ and so $m=h'h^{-1}\implies m\in H$. Thus, $mh\notin H\implies mh=m\implies h=e$. This contradicts our choice of $h$. The conclusion follows.