1. ## 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!

2. 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 $\displaystyle n-1\mid n$ ?

Tonio

3. 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?

4. 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

5. 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?

6. 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 $\displaystyle G-H=\{m\}$. 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.