Using the Second Isomorphism (Diamond Isomorphism) Theorem

[DeldotB]

Homework Statement

Good day all,

Im completely stumped on how to show this:

|AN|=(|A||N|/A intersect N|)

Here: A and N are subgroups in G and N is a normal subgroup.
I denote the order on N by |N|

Homework Equations

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Second Isomorphism Theorem

The Attempt at a Solution

Well, I know know I am supposed to use the theorem, but I have no idea where to start!
I don't seem how to relate what the theorem says to orders of sets...

Any help would be greatly appreciated!

 

[andrewkirk]

First, to be able to use the 2nd isomorphism theorem at all, you need to prove that AN is a subgroup. Has that been given to you as a theorem? If not, can you prove it (it's quite easy)?

Next note that the problem statement only makes sense if A and N are finite.

With those out of the way, the second isomorphism theorem says:

$$AN / N\cong A/A\cap N$$
from which it follows that
$$|AN / N|=| A/A\cap N|$$
You have been asked to prove

$$\frac{|AN|}{|N|}=\frac{|A|}{|A\cap N|}$$

So if you can prove that, for any finite group $G$ and normal subgroup $N$,
$$|G/N| =\frac{|G|}{|N|}$$
then you're almost done.

Can you prove that? Think about the size $|gN|$ of each coset $gN$.

 

[DeldotB]

Ahh! Thanks! I didnt realize I could just divide by |N|...I am new to quotient groups and for some reason I havnen't been able to get the hang of them yet. Thanks!

I'll use Lagranges Theorem for the last part

 

[andrewkirk]

Ahh! Thanks! I didnt realize I could just divide by |N|.

You can do that because the equation you've written is just numbers, as everything is enclosed by |.| signs. So no group theory is needed to justify the division.

Actually, the division is only justified if $|N|\neq 0$ but the proposition is untrue otherwise, since the LHS will be 0 and the RHS will be 0/0, which is undefined. So we assume that the examiner meant to stipulate that $|N|\neq 0$.