# Permutation Groups

• Sep 29th 2008, 03:44 PM
Fulger85
Permutation Groups
Q:

Prove that G is a permutation group if and only if αlpha*(beta^-1) is in G for any alpha, beta in G.

Note: G is a non-empty subset of Sn.

The definition of a group is that for every alpha, beta in g:

a) alpha*beta in G
b) alpha^-1 in G

Help me out!!

Thanks!
• Oct 1st 2008, 07:04 AM
gosualite
Quote:

Originally Posted by Fulger85
Q:

Prove that G is a permutation group if and only if αlpha*(beta^-1) is in G for any alpha, beta in G.

Note: G is a non-empty subset of Sn.

The definition of a group is that for every alpha, beta in g:

a) alpha*beta in G
b) alpha^-1 in G

Help me out!!

Thanks!

The forward implication is trivial.

Suppose then that $\alpha\beta^{-1}\in G$ for any $\alpha,\beta \in G$.

Need to show closure under the operation, associativity, that identities exist, and that an identity element exists.

Well since $\alpha \in G$ (it's nonempty remember), the product $\alpha\alpha^{-1}\in G$ by the given conditions. Do you see how this implies G contains the identity element (call it e)?

Since $\alpha$ and $e$ are in G, then the product $e\alpha^{-1}\in G$. Do you see how this implies G contains inverses for each of its elements?

All you have to do now is acknowledge the inheritance of the associativity property and show that the operation is closed in G.

If you pick $\alpha, \beta \in G$ what gentle subtlety do you need, using what you know, so you can show $\alpha\beta \in G$?
• Oct 1st 2008, 04:01 PM
ThePerfectHacker
There is nothing special about permuation groups.
It can be in fact any group.
Just go over the proof by gosualite and note no fact of it being a permuation group is used.
• Oct 4th 2008, 05:14 PM
Fulger85