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

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.

## 2 Answers

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.

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.

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.