# commutator subgroup eep!

• Apr 30th 2009, 05:03 PM
delilahjam
commutator subgroup eep!
here's the prob

Prove if K is a normal subgroup of a group G then $K'\trianglelefteq G$ where $K'=[K,K] =\langle [a, b]| a, b \in K\rangle$.

I have a theorem i feel like i should be able to use but i'm not sure how...this is it
Let $H\leq G$ and let $x, y\in G$ then $H\trianglelefteq G \iff [H,G]=\leq H$

and

if $G'\leq H$ then $H\trianglelefteq G$.

any ideas?
• Apr 30th 2009, 05:26 PM
ThePerfectHacker
Quote:

Originally Posted by delilahjam
here's the prob

Prove if K is a normal subgroup of a group G then $K'\trianglelefteq G$ where $K'=[K,K] =\langle [a, b]| a, b \in K\rangle$.

I have a theorem i feel like i should be able to use but i'm not sure how...this is it
Let $H\leq G$ and let $x, y\in G$ then $H\trianglelefteq G \iff [H,G]=\leq H$

and

if $G'\leq H$ then $H\trianglelefteq G$.

any ideas?

I think that $g[a,b]g^{-1} = [gag^{-1},gbg^{-1}]$.
• Apr 30th 2009, 05:34 PM
delilahjam
Quote:

Originally Posted by ThePerfectHacker
I think that $g[a,b]g^{-1} = [gag^{-1},gbg^{-1}]$.

I agree, but I can't just take a single commutator. The elements of K' are of the form $[a_1,b_1]^{r_1}[a_2,b_2]^{r_2}\cdots[a_n,b_n]^{r_n}$ for some $r_1, r_2\cdots \in \mathbb{Z}$
• Apr 30th 2009, 05:51 PM
NonCommAlg
Quote:

Originally Posted by delilahjam
I agree, but I can't just take a single commutator. The elements of K' are of the form $[a_1,b_1]^{r_1}[a_2,b_2]^{r_2}\cdots[a_n,b_n]^{r_n}$ for some $r_1, r_2\cdots \in \mathbb{Z}$

it's a simple observation that if a subgroup $H$ of a group $G$ is generated by a set of elements, say $X,$ then $H$ is normal in $G$ if and only if $gxg^{-1} \in H,$ for all $g \in G, \ x \in X.$

the reason is that if $h \in H,$ then $h=x_1^{\pm 1} x_2^{\pm 1} \cdots x_n^{\pm 1},$ for some $x_i \in X.$ but then: $ghg^{-1}=(gx_1g^{-1})^{\pm 1}(gx_2 g^{-1})^{\pm 1} \cdots (gx_ng^{-1})^{\pm 1}.$ i think you see my point now!
• Apr 30th 2009, 07:58 PM
ziggychick
awesome thanks!
• Apr 30th 2009, 08:02 PM
delilahjam
actually thanks from me!

i guess we love mathhelpforum here so much we never sign out!