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 $g^{-1}Ng$ when multiply, it still in $g^{-1}Ng$, not in G.

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

and you need to show x*y is in $g^{-1}Ng$

p/s: sorry if my english terrible