Show that a subgroup is normal

Let G be a group. Prove that $\displaystyle N = \langle x^{-1} y^{-1} xy \, \, | \, \, x, y \in G \rangle$ is a normal subgroup of G and $\displaystyle G/N$ is abelian

So starting with the first part...

To show something is a normal subgroup we have to show that $\displaystyle g (x^{-1} y^{-1} xy) g^{-1} \in N$ for some g in G.

I've been trying to rearrange that equation to look something line the definition of N but can't seem to get anywhere...

$\displaystyle g x^{-1} y^{-1} xy g^{-1} \cong g^{-1} x^{-1} y^{-1} xy g = (yxg)^{-1} xyg = g^{-1} (yx)^{-1} (xy)g = ...$

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Re: Show that a subgroup is normal

Hi,

Here's a proof together with some convenient notation in groups:

http://i59.tinypic.com/5b31av.png

Re: Show that a subgroup is normal

This is what you are missing:

Suppose we have $xyx^{-1}y^{1}$ and we take $g(xyx^{-1}y^{-1})g^{-1}$.

We can re-write this as:

$g(xyx^{-1}y^{-1})g^{-1} = (gxg^{-1})(gyg^{-1})(gx^{-1}g^{-1})(gy^{-1}g^{-1})$ (the $g$'s in the middle cancel)

$= (gxg^{-1})(gyg^{-1})((g^{-1})^{-1}x^{-1}g^{-1})((g^{-1})^{-1}y^{-1}g^{-1})$ (since $g = (g^{-1})^{-1}$)

$= (gxg^{-1})(gyg^{-1})(gxg^{-1})^{-1}(gyg^{-1})^{-1} \in N$.

There is one small thing you are overlooking:

$N$ isn't JUST elements of the form $xyx^{-1}y^{-1}$, it's all finite PRODUCTS of that form: $n_1n_2\cdots n_k$, where $n_j = x_jy_jx_j^{-1}y_j^{-1}$ for each $j$.

HOWEVER, it isn't hard to show that:

$g(n_1n_2\cdots n_k)g^{-1} = (gn_1g^{-1})(gn_2g^{-1})\cdots(gn_kg^{-1})$ (this is the same trick we used before), and since each term $gn_jg^{-1} \in N$ (this is what we proved above) so is their product (since $N$ is closed under multiplication).

The notation $gxg^{-1}$ is unwieldy, so algebraists have come up with a better form:

$x^g = g^{-1}xg$ (why do we put the "inverse" first? so that we have $(x^g)^h = x^{gh}$ otherwise we'd have to "reverse the order").

Along a similar vein, we define $[x,y] = x^{-1}y^{-1}xy$ (this gives us the same commutators as before, because elements and inverses are in one-to-one correspondence).

As johng's post shows, we have:

$[x,y]^g = [x^g,y^g]$ which is "algebraically neat".

Here is a useful theorem:

$G$ abelian iff $N = \{e\}$. This is why we use commutators, they tell us "how far away from abelian" $G$ is.

To see that $G/N$ is abelian observe:

$(Ny)(Nx) = N(yx) = (N)(N(yx)) = (N(xyx^{-1}y^{-1}))(N(yx))$ (since $xyx^{-1}y^{-1} \in N$)

$= N(xyx^{-1}y^{-1}yx) = N(xyx^{-1}x) = Nxy = (Nx)(Ny)$.

In fact, we can actually say MORE:

Suppose we have a homomorphism $\phi:G \to A$, where $A$ is abelian. Then $N \subseteq\text{ker }\phi$ (prove this! it isn't hard), so $N$ is the SMALLEST normal subgroup of $G$ that makes $G/N$ abelian.