Is g^-1Ng a Subgroup of G? Proving Invariance in Group Theory

[Juanriq]

Salutations all, just stuck with the starting step, I want to see if I can take it from there.

Homework Statement

Let G be a group and let N be a subgroup of G. Prove that the set g^{-1}Ng is a subgroup of G.

The Attempt at a Solution

Well, I'm going to have to show that g^{-1}Ng is closed and contains an inverse. Do I start by saying that g \in G and n \in N, therefore n \in G as well as g^{-1} \in G. The fact that G is a group means that combining these terms under the operation will still fall in G because it is closed. Also, for inverses, the element n^{-1}\in G so I can take g^{-1}n^{-1}g as the inverse?

Thanks in advance!

 

[annoymage]

if you want to show their element is closed, you have to show any two element in <br /> g^{-1}Ng <br /> when multiply, it still in <br /> g^{-1}Ng <br />, not in G.

so if x and y is in <br /> g^{-1}Ng <br />, what can you say about x and y??

and you need to show x*y is in <br /> g^{-1}Ng <br />

p/s: sorry if my english terrible