# Thread: A Few Normal SUbgroup Questions

1. ## A Few Normal SUbgroup Questions

1) Let H normal to G and K normal to G. Show that H intersection K is normal to G.

2) Let H normal to G and K normal to G, and assume that H intersection K = {e}. Show that if x is in H and y is in K, then xy = yx.

3) Show that (Q,+) / (Z,+) is an infinite group every element of which has finite order.

4) Let G be a group and let H be a subgroup of index 2. Show that for every a in G, a^2 is in H.

2. 1) If an element, x, is in $\displaystyle H \cap K$ then it is in H and K, since all elements in H and K have their conjugates in H and K, all the conjugates of x will be in $\displaystyle G \cap K$

2) Let $\displaystyle x \in H$, $\displaystyle y \in K$ Consider $\displaystyle \underbrace{(y^{-1})x(y^{-1})^{-1}}_{\in H} x^{-1} \in H$ but $\displaystyle (y^{-1})\underbrace{x(y^{-1})^{-1} x^{-1}}_{\in K} \in K$so $\displaystyle y^{-1}xyx^{-1} = e \Rightarrow xy = yx$

3) This is the group of elements $\displaystyle \frac{m}{n} + \mathbb{Z}$, so add each element to itself n (which is finite and specific to each element) times

3. I still don't get 3. I thought it would be a rational number over some integer? Not m/n + Z?

And does anyone know #4?

4. Originally Posted by Janu42
I still don't get 3. I thought it would be a rational number over some integer? Not m/n + Z?

And does anyone know #4?
Yes! You should remember that if H is a subgroup of index 2, it must be normal in G. Thus the factor group G/H={H,aH} has two elements, and therefore $\displaystyle a^{2}H=(aH)^{2}=H$, that is, $\displaystyle a^{2}\in H$.

5. OK Thanks.

One more: How do I show the special linear group is normal to the general linear group?

And how do I show that if H is the subgroup of G with 2 x2 matrices where c = 0 (ad does not equal 0), then H is not a normal subgroup of G? I'm assuming if I know how to do the first one I can get this.

EDIT: I saw something about using the special linear group as a kernel or something, but I don't think we know that yet. So if there's a way to use the basic definition of normal subgroups for this, that works.

6. Originally Posted by Janu42
OK Thanks.

One more: How do I show the special linear group is normal to the general linear group?

And how do I show that if H is the subgroup of G with 2 x2 matrices where c = 0 (ad does not equal 0), then H is not a normal subgroup of G? I'm assuming if I know how to do the first one I can get this.

EDIT: I saw something about using the special linear group as a kernel or something, but I don't think we know that yet. So if there's a way to use the basic definition of normal subgroups for this, that works.
determinant is a homomorphism from the general linear group on a field to the multiplicative group of that field. the kernel of this homomorphism is the set of all matrices with determinant 1, which

is exactly the definition of the special linear group. for the second part of your problem, let $\displaystyle g=\begin{pmatrix}1 & 0 \\ 1 & 1 \end{pmatrix}, \ \ h=\begin{pmatrix}1 & 1 \\ 0 & 1 \end{pmatrix} \in H.$ show that for all $\displaystyle x \in H: \ gh \neq xg.$

7. Is this the only way to do it? Is there a way to use the definition of normal subgroup simply?

Or is this simpler than I'm making it seem? I don't really get the idea of kernel.