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Suppose G is an finite group. Let N be a normal subgroup of G and A be an arbitraty subgroup. Prove that $$\mid AN\mid=\frac{\mid A\mid\mid N\mid}{\mid A\cap N\mid}$$

My initial approach is that I should somehow invoke one of the isomorphism theorems that involves quotient group and then apply another theorem to grasp about its size. Yet I've no idea how to actually approach it, especially with the given condition that $N$ is normal.

Allen Ai
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1 Answers1

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Hint: First show that $AN$ is a subgroup of $G$, then consider the map $\phi : A \to \frac{AN}{N}$ and use first iso theorem.

AnotherPerson
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  • I can see (and show) that AN is a subgroup of G. Could you please elaborate on the latter part? Thanks! – Allen Ai Oct 26 '16 at 07:43
  • So what is the identity element of $\frac{AN}{N}$, and what elements get sent to it? – AnotherPerson Oct 26 '16 at 07:45
  • So $\frac{AN}{N}$ is all the left cosets of $N$ in $AN$: ${ aN : a\in AN}$. Then if $a\in A$, $a\rightarrow aN$, otherwise if $a\in N$ then $a\rightarrow e$? – Allen Ai Oct 26 '16 at 08:08
  • Or should I think about it this way: we know that $\phi (A)\cong\frac{AN}{N}\cong\frac{A}{A\cap N}$ (arrived with Diamond Isomorphism Theorem), now since $N\subset AN$ and $(A\cap N)\subset A$ by Lagrange's theorem $\frac{\mid AN\mid}{\mid N\mid}$ is the the number of left cosets of $N$ in $AN$ and similarly $\frac{\mid A\mid}{\mid A\cap N\mid}$ the number of left cosets in $A$. Then this way because of the isomorphism between these two groups, $\frac{\mid AN\mid}{\mid N\mid} = \frac{\mid A\mid}{\mid A\cap N\mid}$. With simple algebra this would give me the desired statement? – Allen Ai Oct 26 '16 at 08:25
  • N is the identity of $\frac{AN}{N}$. – AnotherPerson Oct 26 '16 at 16:05