# Let G be a group, let H < G, and let x E G. We use the notation xHx1 to denote theset of elements {xhx1h E H}. (xHx is called a conjugate of H.)Prove that xHx-1 is a subgroup of G(a)If H is finite, then how are |H| and |xHx(b)(c)related?Prove that H is isomorphic to xHx1

Question
Step 1

In a group G, two element g and h are called conjugate when h = x g x-1 for some x G

For an element g of a group G, its conjugacy class is the set of elements conjugate to it:

{xgx−1 : x ∈ G}.

Part (a)          The set xHx-1 is a subgroup of G.

Proof: Since H is a subgroup, we know that H is nonempty, and therefore xHx-1 is nonempty.

Next,

let x h1 x-1 and x h2 x-1 be elements of xHx-1, where h1, h2 ∈ H.

Since H is a subgroup, we know that h1, h2 ∈ H, and therefore (x h1 x-1)( x h2 x-1  ) = xh1h2 x-1 is also an element of xHx-1.

Finally, let xhx-1 be any element of xHx-1 , where h ∈ H.

Since H is a subgroup, we know that h-1 ∈ H,

and therefore (x h1 x-1)-1= x h-1 x-1  is an element of xHx-1

Step 2

Part (b):

If H is a finite group, then each conjugacy class in H has size dividing |H|

Step 3

An isomorphism form H to xHx-1 is one-one mapping from H onto xHx-1 and preserve the group operation and xHx-1 is homomorphism:

First prove that...

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