3
$\begingroup$

Suppose G is a finite group. Let N be a normal subgroup of G and A an arbitrary subgroup. Verify that $|AN|=\frac{|A|*|N|}{|A \cap N|}$.

I think this is fairly intuitive. So $AN=\{an | a \in A, n \in N\}$. If they are disjoint, then this makes sense because it will just be an element from $A$ times an element of $N$ giving $|A|*|N|$. However, I am unsure how to prove this or why the fact that $N$ is normal comes in. So my question is, how do I prove this statement?

  • 2
    Indeed, $N$ being normal will not come into play as it is not necessary for this claim (that would just guarantee for the product to be a subgroup rather than just a subset). Write the product as a union of cosets of one of the subgroups to see why the order is as it is. – 2017-02-21

2 Answers 2

2

Prove that $AN=\{ax:a\in A, x\in N\}$ is a subgroup of $G$ (this depends on $N$ being normal, otherwise it's not generally true).

Now consider the map $$ f\colon A\to AN/N,\qquad f(a)=aN $$ This is a surjective group homomorphism, with kernel $A\cap N$. The isomorphism theorem tells you that $$ A/(A\cap N)\cong AN/N $$ Therefore, by Lagrange, $$ \frac{|A|}{|A\cap N|}=\frac{|AN|}{|N|} $$

2

Let $w \in AN$ then there are an $a \in A$ and a $n \in N$ such that $w=an$. This representation of $w$ isn't necessarily unique but it always has the form $w=\underbrace{ad}_{\text{=a'}} \underbrace{d^{-1}n}_{\text{=n'}}=an $ since $A$ and $N$ are subgroups it follows that $d \in A \cap N$. That means that for each $w \in AN$ there are $|A \cap N|$ many representations. The statement follows as claimed.