Thread: 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!

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 ?

Tonio

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?

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

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?

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 . Now, clearly . Now, since we see that and so there exists some . Clearly but since if it were we'd have that and so . Thus, . This contradicts our choice of . The conclusion follows.