# Definition of normal subgroup

1. Feb 6, 2007

### quasar987

My professor of topology gave us a quick overview of the group theory results we will be needing later and among the things he said, is that a normal subgroup of a group G is a subgroup H such that for all x in G, xHx^{-1}=H.

Is this correct? The wiki article seems to indicate that equality btw xHx^{-1} and H is unnecessary, but rather that inclusion is sufficient.

2. Feb 7, 2007

### matt grime

They are easily shown to be equivalent.

3. Feb 7, 2007

### quasar987

Mmmh... So for all x in G, $xHx^{-1}\subset H \Leftrightarrow xHx^{-1}= H$.

For a fixed x, suppose there exists h' in H such that there are no h in H with h'=xhx^{-1}. This is nonsense since h=x^{-1}h'x does the trick.

Last edited: Feb 7, 2007
4. Feb 7, 2007

### StatusX

But then how do you know h is in H? And I don't see what this has to do with your original question.

5. Feb 7, 2007

### quasar987

My original question was (essentially) Are the two definitions of normal subgroup "H is is normal subgroup of G if for all x in G, $xHx^{-1}\subset H$" and "H is is normal subgroup of G if for all x in G, $xHx^{-1}= H$" equivalent?

Then matt grime said "Yes". Then I tried to prove this assertion, so I said the following: "To show they are equivalent, I only need to show that for all x in G, $xHx^{-1}\subset H \Rightarrow xHx^{-1}= H$ since the converse is immediate."

So let's proceed by contradiction by supposing that for a certain fixed x in G, there exists an h' in H such that there are no h in H with h' = xhx^{-1}. This is automatically seen to be an absurdity since the element h of G defined by h=x^{-1}h'x is such that h' = xhx^{-1} and it is in H because by hypothesis, h' is in H and $xHx^{-1}\subset H$, i.e. $h=x^{-1}h'x\in H$.

I had to write it all in large to convince myself fully that it is correct, and still seems so to me. Do you still have an objection StatusX?

Last edited: Feb 7, 2007
6. Feb 7, 2007

### StatusX

That's clearer than your last post, but still a little hard to follow, with an unnecessary proof by contradiction. The proof is easy:

Assume for all x in G, xHx^-1 is a subset of H. Then for all h in H, we can define h'=x^-1hx, which is in H, and so h=xh'x^-1, ie, H is a subset of xHx^-1.