# Diamond Isomorphism Theorem clarification. I'm good on (a) and (b). I am confused about (c)

AN/N would be the the set of all left cosets of N in AN right?

An element of AN/N would look like this: (an)N

Why is this isomorphic to $\varphi$(A)?

And then why is this isomorphic to A/(A$\cap$N)?

• Suppose $\varphi:G\to G/N$. Then not only is $AN/N$ isomorphic to $\varphi(A)$, it is equal to it, because $\varphi(\varphi^{-1}(\varphi(A)))=\varphi(A)$ is a set-theoretic fact. As for your last question: guess what the isomorphism $AN/N\to A/(A\cap N)$ should be, then prove it is one. – anon Nov 15 '15 at 6:07
• Note that $(an)N = aN$. – David Wheeler Nov 15 '15 at 6:08
• Okay that's what I didn't realize. Why don't they just express it as A/N? And I'm unsure about that last isomorphism. Any hints? I know (A$\cap$N) is the set of elements of N in A. But why is AN/N bijective with A/ (A$\cap$N). – user139985 Nov 15 '15 at 6:16
• @user139985: $A/N$ would not be correct because $N$ is not necessarily contained in $A$. We need to look for a larger subgroup which contains both $A$ and $N$. And indeed, $AN$ is the smallest subgroup which contains both. – Bungo Nov 16 '15 at 4:30

(1) First statement: $\varphi^{-1}(\varphi(A))=AN$. What this says?

(2) We have $\varphi\colon G\rightarrow G'$, and $\varphi(A)$ is a subgroup of $G'$. Its a subgroup of $G'$; from which largest subgroup of $G$ it is coming (via $\varphi$?) It is precisely $\varphi^{-1}(\varphi(A))$, which is $AN$.

(3) Thus, $\varphi(A)$ is coming from $AN$ via $\varphi$, and not from other elements in $G$. So to consider (c), forget $G$ and consider $\varphi$ only on $AN$.

(4) From $AN$ to $\varphi(A)$ we can define a natural map: $an\mapsto \varphi(a)$. This is clearly surjective [$\varphi(AN)=\varphi(A)\varphi(N)=\varphi(A).1=\varphi(A)$].

(5) What is kernel of above map? If $an\mapsto 1$, i.e. $\varphi(a)=1$ then $a\in \ker\varphi=N$, and conversely, every element of $N$ goes to 1 under map in (4).

(6) So in (4) we have a surjective homomorphism from $AN$ to $\varphi(A)$, whose kernel is $N$. Does this gives some isomorphism in (c)?

(7) We do almost same game with slight modification: consider map from $A$ to $\varphi(A)$ by $a\mapsto \varphi(a)$. [We take here $A$ instead of $AN$ as in (4)].

(8) Again this is surjective and $a\mapsto 1$ if and only if $\varphi(a)=1$ i.e. if and only if $a\in\ker\varphi=N$ i.e. if and only if $a$ is in $A$ as well as $N$ ($a\in A\cap N$).

(9) So the map in (7) is surjective from $A$ to $\varphi(A)$ with kernel $A\cap N$. Can you complete now?