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!

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- Mar 3rd 2010, 04:26 AMmarchesabstract 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! - Mar 3rd 2010, 04:53 AMtonio
- Mar 3rd 2010, 05:34 AMmarches
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?

- Mar 3rd 2010, 06:17 AMtonio
- Mar 3rd 2010, 06:27 AMmarches
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?

- Mar 3rd 2010, 07:23 AMDrexel28
We have that $\displaystyle G-H=\{m\}$. Now, clearly $\displaystyle m\ne e$. Now, since $\displaystyle n>2$ we see that $\displaystyle |H|>1$ and so there exists some $\displaystyle h\in H-\{e\}$. Clearly $\displaystyle mh\in G$ but $\displaystyle mh\notin H$ since if it were we'd have that $\displaystyle mh=h',\text{ }h'\in H$ and so $\displaystyle m=h'h^{-1}\implies m\in H$. Thus, $\displaystyle mh\notin H\implies mh=m\implies h=e$. This contradicts our choice of $\displaystyle h$. The conclusion follows.