The quotient of a quotient group by another quotient group

Question

Let $G$, $M$, and $N$ be the cyclic groups given by $$ G \colon= \left\langle a \colon a^{12} = e \right\rangle = \left\{ e = a^0, a, a^2, \ldots, a^{11} \right\}, $$ $$ M \colon= \left\langle a^2 \right\rangle = \left\{ e, a^2, a^4, a^6, a^8, a^{10} \right\}, $$ and $$ N \colon= \left\langle a^6 \right\rangle = \left\{ e, a^6 \right\}. $$

Then of course $M$ is a normal subgroup of $G$, and of course $N$ is a normal subgroup of both $G$ and $M$.

Thus the quotient groups $G/M$, $G/N$, and $M/N$ are well-defined.

In fact, we have $$ \begin{align} G/M &= \left\{ M, aM, a^3M, a^5M, a^7M, a^9M, a^{11}M \right\} \\ &= \left\{ M, aM \right\} \\ &= \left\{ \, \left\{ e, a^2, a^4, a^6, a^8, a^{10} \right\}, \, \left\{ a, a^3, a^5, a^7, a^9, a^{11} \right\} \, \right\}, \end{align} $$ $$ \begin{align} G/N &= \left\{ N, aN, a^2N, a^3N, a^4N, a^5N, a^7N, a^8N, a^9N, a^{10}N, a^{11}N \right\} \\ &= \left\{ N, aN, a^2N, a^3N, a^4N, a^5N \right\} \\ &= \left\{ \, \left\{ e, a^6 \right\}, \, \left\{ a, a^7 \right\}, \, \left\{ a^2, a^8 \right\}, \, \left\{ a^3, a^9 \right\}, \, \left\{ a^4, a^{10} \right\}, \, \left\{ a^5, a^{11} \right\} \, \right\}, \end{align} $$ and $$ \begin{align} M/N &= \left\{ N, a^2N, a^4N, a^8 N, a^{10}N \right\} \\ &= \left\{ N, a^2N, a^4 N \right\} \\ &= \left\{ \, \left\{ e, a^6 \right\}, \, \left\{ a^2, a^8 \right\}, \, \left\{ a^4, a^{10} \right\} \, \right\}. \end{align} $$

Furthermore, as $M$ is a normal subgroup of $G$, so the quotient group $M/N$ is a normal subgroup of the quotient group $G/N$.

Therefore we can consider the quotient group $(G/N)/(M/N)$.

Now my question is, is the following construction valid?

We first note that $$ \begin{align} \left\{ a, a^7 \right\} (M/N) &= \left\{ a, a^7 \right\} \left\{ \, \left\{ e, a^6 \right\}, \, \left\{ a^2, a^8 \right\}, \, \left\{ a^4, a^{10} \right\} \, \right\} \\ &= \left\{ \, \left\{ a, a^7 \right\} \left\{ e, a^6 \right\}, \, \left\{ a, a^7 \right\} \left\{ a^2, a^8 \right\}, \, \left\{ a, a^7 \right\} \left\{ a^4, a^{10} \right\} \, \right\} \\ &= \left\{ \, \left\{ a, a^7 \right\}, \, \left\{ a^3, a^9 \right\}, \, \left\{ a^5, a^{11} \right\} \, \right\}. \end{align} $$ And also $$ \begin{align} \left\{ a^3, a^9 \right\} (M/N) &= \left\{ a^3, a^9 \right\} \left\{ \, \left\{ e, a^6 \right\}, \, \left\{ a^2, a^8 \right\}, \, \left\{ a^4, a^{10} \right\} \, \right\} \\ &= \left\{ \, \left\{ a^3, a^9 \right\} \left\{ e, a^6 \right\}, \, \left\{ a^3, a^9 \right\} \left\{ a^2, a^8 \right\}, \, \left\{ a^3, a^9 \right\} \left\{ a^4, a^{10} \right\} \, \right\} \\ &= \left\{ \, \left\{ a^3, a^9 \right\}, \, \left\{ a^5, a^{11} \right\}, \, \left\{ a, a^7 \right\} \, \right\} \\ &= \left\{ \, \left\{ a, a^7 \right\}, \, \left\{ a^3, a^9 \right\}, \, \left\{ a^5, a^{11} \right\} \, \right\}. \end{align} $$ Thus we have shown that $$ \left\{ a, a^7 \right\} (M/N) = \left\{ a^3, a^9 \right\} (M/N). $$ Similarly we can show that $$ \left\{ a, a^7 \right\} (M/N) = \left\{ a^5, a^{11} \right\} (M/N). $$

Using the calculations above, we find that $$ \begin{align} (G/N)/(M/N) &= \left\{ \ M/N, \ \left\{ a, a^7 \right\} (M/N) , \ \left\{ a^3, a^9 \right\}(M/N), \ \left\{ a^5, a^{11} \right\} (M/N) \ \right\} \\ &= \left\{ \ M/N, \ \left\{ a, a^7 \right\} (M/N) \ \right\} \\ &= \left\{ \ \left\{ \, \left\{ e, a^6 \right\}, \, \left\{ a^2, a^8 \right\}, \, \left\{ a^4, a^{10} \right\} \, \right\}, \ \left\{ a, a^7 \right\} \left\{ \, \left\{ e, a^6 \right\}, \, \left\{ a^2, a^8 \right\}, \, \left\{ a^4, a^{10} \right\} \, \right\} \ \right\} \\ &= \left\{ \ \left\{ \, \left\{ e, a^6 \right\}, \, \left\{ a^2, a^8 \right\}, \, \left\{ a^4, a^{10} \right\} \, \right\}, \ \left\{ \, \left\{ a, a^7 \right\} \left\{ e, a^6 \right\}, \, \left\{ a, a^7 \right\} \left\{ a^2, a^8 \right\}, \, \left\{ a, a^7 \right\} \left\{ a^4, a^{10} \right\} \, \right\} \ \right\} \\ &= \left\{ \ \left\{ \, \left\{ e, a^6 \right\}, \, \left\{ a^2, a^8 \right\}, \, \left\{ a^4, a^{10} \right\} \, \right\}, \ \left\{ \, \left\{ a, a^7 \right\}, \, \left\{ a^3, a^9 \right\}, \, \left\{ a^5, a^{11} \right\} \, \right\} \ \right\}. \end{align} $$

Of course the quotient group $(G/N)/(M/N)$ is isomorphic to the quotient group $G/M$.

Is my construction correct?

Is each and every detail of my calculation above correct? Or, have I made any errors or logical / mathematical mistakes?

Last but not the leat, is my typesetting logically correct and clear enough as well? Or, is there a better way of presenting the work above?

Answer 1

Not only is your conclusion correct, you can probably find it stated as a theorem (possibly with general proof) in your favorite group-theory textbook [e.g it is in my copy of Fraleigh 3rd edition].

Answer 2

Theorem: Let N,K be normal subgroups of G with $N\subseteq K\subseteq G$, then (1) N is a normal subgroup of K,(2) $K/N$ is a normal subgroup of $G/N$, and (3) $(G/N)/(K/N)\cong G/K$.

Proof:

(1): Since N is a normal subgroup of G, $N=gNg^{-1}$ for all $g\in G$. In particular, $N=kNk^{-1}$ for all $k\in K$ since $K\subseteq G$, which means that N is a normal subgroup of K.

(2): Define a surjective homomorphism $f: G\rightarrow f(G)$ such that $N=\ker f$. Then $f_K$, the restriction of $f$ onto K, also has kernel N. By the first isomorphism theorem, $G/N\cong f(G), K/N\cong f(K).$ Since f is a homomorphism and $K=gKg^{-1}$ for every $g\in G$, we have $f(K)=f(g)f(K)f(g)^{-1}$ which implies $f(K)$ is normal in $f(G)$. Therefore, $K/N$ is normal in $G/N$.

(3) Define a surjective homomorphism $\pi: G\rightarrow G/K$ canonically by $g \mapsto gK$, then it has kernel K and induces a homomorphism $ \phi: G/N\rightarrow G/K$ with $\phi(gN)=\pi(g)=gK$. Since N is normal to K, the elements in the same coset $gN$ are mapped into the same coset $gK$. Thus, $\phi$ is well-defined. Now, $\phi$ is surjective and $\ker \phi=K/N$, hence by the first isomorphism theorem we have $(G/N)/(K/N)\cong G/K$.

I probably skipped over some details, but my point is that the observation you made is actually something that works for more general objects.