What does it mean for a group to be a quotient of another and how does it imply isomorphism?

Question

I was reading a proof of the following proposition and had one small doubt about the proof:

Proposition: Let $H$ be a subgroup of $G,$ and $\mathcal{A}$ a $(\text {sub})$ normal series in $G .$ Then the series $$ \mathcal{A}_{H}: E=A_{0} \cap H \subset A_{1} \cap H \subset \ldots \subset A_{l} \cap H=H $$ is a (sub) normal series in $H$ having factors which are isomorphic to subgroups of the factors of $\mathcal{A} .$ If $H \lhd G,$ then the series $$ \overline{\mathcal{A}}: E=A_{0} H / H \subset A_{1} H / H \subset \ldots \subset A_{l} H / H=G / H $$ is a (sub)normal series for $G / H$ having factors which are isomorphic to quotients of the factors of $\mathcal{A}$

Proof: It is obvious that $\mathcal{A}_{H}$ is a (sub)normal series in $H .$ The homomorphism $A_{i+1} \cap H \rightarrow A_{i+1} / A_{i}$ obtained by restricting the natural homomorphism $A_{i+1} \rightarrow A_{i+1} / A_{i}$ to $A_{i+1} \cap H$ has kernel $A_{i} \cap\left(A_{i+1} \cap H\right)=A_{i} \cap H$. Therefore the factor $\left(A_{i+1} \cap H\right) /\left(A_{i} \cap H\right)$ of $\mathcal{A}_{H}$ is isomorphic to a subgroup of the factor $A_{i+1} / A_{i}$ of $\mathcal{A} .$ Suppose now that $H \lhd G .$ Then $A_{i} H \lhd A_{i+1} H$ hence $A_{i} H / H \lhd A_{i+1} H / H,$ and the quotient is isomorphic to $$ A_{i+1} H / A_{i} H \simeq A_{i+1} / A_{i}\left(H \cap A_{i+1}\right) $$ which is a quotient of $A_{i+1} / A_{i} .$ If $A_{i}\lhd G,$ then $A_{i} H / H \lhd G / H.$

I have understood the entire proof except what "which is a quotient of $A_{i+1} / A_{i}$" implies and how that helps establish an isomorphism with the series $\mathcal{A}$.

Please help, any will be very much appreciated. Thank you!

Luthar, I. S., Algebra. Volume 1: Groups, New Delhi: Narosa Publishing House. xxxvi, 442 p. (1996). ZBL0943.20001.

Answer 1

The goal seems to be to prove, for arbitrary $0\leq i< l$, that the quotient $(A_{i+1}H/H)/(A_iH/H)$ is isomorphic to a quotient of $A_{i+1}/A_i$ (because that is the form of the factors of $\mathcal{A}$).

The author used the fact that $(A_{i+1}H/H)/(A_iH/H)$ is isomorphic to $A_{i+1}H/A_iH$, and then further reduced this factor via an isomorphism to $A_{i+1}/(A_iH)\cap A_{i+1}=A_{i+1}/A_i(H\cap A_{i+1})$ (it seems like you understand this part).

So far, we have $(A_{i+1}H/H)/(A_iH/H)\cong A_{i+1}/A_i(H\cap A_{i+1})$. Now we just need to show that $A_{i+1}/A_i(H\cap A_{i+1})$ is isomorphic to some quotient of $A_{i+1}/A_i$.

Consider the function $f:A_{i+1}/A_i\rightarrow A_{i+1}/A_i(H\cap A_{i+1})$ defined as $h(A_i\cdot a)=A_i(H\cap A_{i+1})\cdot a$ (where $A_i\cdot a$ is the coset of $A_i$ with some $a\in A_{i+1}$ and similarly $A_i(H\cap A_{i+1})\cdot a$ is the coset of $A_i(H\cap A_{i+1})$ with $a$).

With a tiny bit of work, one can show that $f$ is a well-defined surjective homomorphism.

Then (by the first isomorphism theorem) $A_{i+1}/A_i(H\cap A_{i+1})$ will be isomorphic to a quotient group of $A_{i+1}/A_i$. Since $(A_{i+1}H/H)/(A_iH/H)\cong A_{i+1}/A_i(H\cap A_{i+1})$, our proof will be complete.

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Proof that $f$ is a well-defined surjective homomorphism:

It is obvious that $f$ maps into $A_{i+1}/A_i(H\cap A_{i+1})$. If for any pair $a,b\in A_{i+1}$ we have $A_i\cdot a=A_i\cdot b$, then $ab^{-1}\in A_i$ so $ab^{-1}\in A_i(H\cap A_{i+1})$ giving us $A_i(H\cap A_{i+1})\cdot a=A_i(H\cap A_{i+1})\cdot b$, i.e. $f(A_i\cdot a)=f(A_i\cdot b)$. So $f$ is well-defined.

Also, for any $A_i\cdot a,\;A_i\cdot b\in A_{i+1}/A_i$, we have $$f((A_i\cdot a)(A_i\cdot b))=f(A\cdot (ab))=A_i(H\cap A_{i+1})\cdot (ab)$$$$=(A_i(H\cap A_{i+1})\cdot a)(A_i(H\cap A_{i+1})\cdot b)=f(A_i\cdot a)f(A_i\cdot b)$$

So $f$ is a true homomorphism.

Finally, let $A_i(H\cap A_{i+1})\cdot a$ be any element $\in A_{i+1}/A_i(H\cap A_{i+1})$. Then $a\in A_{i+1}$ so $A_i\cdot a\in A_{i+1}/A_i$. This means we can say that $A_i(H\cap A_{i+1})\cdot a=f(A_i\cdot a)$, making $f$ surjective.

We conclude that $f:A_{i+1}/A_i\rightarrow A_{i+1}/A_i(H\cap A_{i+1})$ is a well-defined surjective homomorphism.

Answer 2

I am going to address your title question, and let you apply it to the specific example.

You should be familiar with the first isomorphism theorem. Well I think it actually answers your question.

Any time you have a quotient of one group $G$ by a normal subgroup $N$, there's the canonical homomorphism $h:G\mapsto G/N$ given by $h(g)=gN$.

This quotient satisfies a universal property: for any homomorphism $\phi$ from another group to $G/N$, it "factors through" $h$.

I believe your question is not a particularly hard one. The upshot is that you can form the quotient whenever you have a normal subgroup.