abstract algebra proof

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March 3rd 2010, 04:26 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, 04:53 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, 05:34 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, 06:17 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, 06:27 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, 07:23 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.