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From this wikipedia's page, I have found these important theorems. I would like to give it a try. Since I am self-learning Mathematical Analysis without teacher or tutor, it will be great if someone helps me check it out. Thank you for your help!

Third isomorphism theorem on groups

Let $G$ be a group and $N$ a normal subgroup of $G$.

  1. If $K$ is a subgroup of $G$ such that $N \subseteq K \subseteq G$, then $K/N$ is a subgroup of $G/N$.

  2. Every subgroup of $G/N$ is of the form $K/N$, for some subgroup $K$ of $G$ such that $N \subseteq K \subseteq G$.

  3. If $K$ is a normal subgroup of $G$ such that $N \subseteq K \subseteq G$, then $K/N$ is a normal subgroup of $G/N$.

  4. Every normal subgroup of $G/N$ is of the form $K/N$, for some normal subgroup $K$ of $G$ such that $N \subseteq K \subseteq G$.

  5. If $K$ is a normal subgroup of $G$ such that $N \subseteq K \subseteq G$, then the quotient group $(G/N)/(K/N) \cong G/K$.

My attempt:

Clearly, $K/N,G/N$ are groups and $K/N \subseteq G/N$. So $K/N$ is a subgroup of $G/N$. Assertion (1.) then follows.

Let $H$ be a subgroup of $G/N$ and $K = \{g \in G \mid gN \in H\}$. If $g_1, g_2 \in K$ then $g_1N \in H,g_2N \in H$. So $(g_1N)(g_2N) \in H$ or $(g_1g_2)N \in H$. Hence $g_1g_2 \in K$ and thus $K$ is closed. If $g \in K$ then $gN \in H$ and thus $g^{-1}N = Ng^{-1} = N^{-1}g^{-1} = (gN)^{-1} \in H$. Hence $g^{-1} \in K$. So $K$ is a group. Clearly, $H = K/N$. Since $N$ is the identity element of $G/N$ and $H$ a subgroup of $G/N$, $N \in H$. We have $n \in N \implies nN=N \in H \implies$ $n \in K \implies N \subseteq K$. It follows that $H = K/N$. Assertion (2.) then follows.

Because of (1.), it suffices to prove that $K/N$ is normal in $G/N$. For all $kN \in K/N,gN \in G/N$, $(gN)(kN)(gN)^{-1} = gNNkN^{-1}g^{-1} = gNkNg^{-1} =$ $gNNkg^{-1} = gNkg^{-1} = gkNg^{-1} =$ $gkg^{-1}N \in K/N.$ Assertion (3.) then follows.

Because of (2.), it suffices to prove that $K$ is normal in $G$. Since $K/N$ is normal in $G/N$, $\forall (kN \in K/N, gN \in G/N): (kN)(gN)(kN)^{-1} \in K/N$. On the other hand, $(kN)(gN)(kN)^{-1} \in K/N \iff kNNgN^{-1}k^{-1} \in K/N \iff kNgNk^{-1} \in K/N$ $\iff$ $kgNNk^{-1} \in K/N \iff kgNk^{-1} \in K/N \iff kgk^{-1}N \in K/N \iff$ $kgk^{-1} \in K$. This implies $\forall (k \in K, g\in G): kgk^{-1} \in K$ and thus $K$ is normal in G. Assertion (4.) then follows.

Consider

$$\begin {array}{lrcl} \psi : & G/N & \longrightarrow & G/K\\ & gN & \longmapsto & gK \end{array}$$

Clearly, $\psi$ is surjective. We have $$\psi((g_1N)(g_2N)) = \psi(g_1g_2N) = g_1g_2K = (g_1K)(g_2K) =\psi(g_1N) \psi(g_2N)$$ Hence $\psi$ is a homomorphism.

$\ker (\psi) = \{gN \in G/N \mid \psi(gN) = K\} = \{gN \in G/N \mid gK = K\}$. On the other hand, $gK=K \iff g \in K$. Hence $\ker (\psi) =K/N$.

By the first isomorphism theorem on groups, we have $$(G/N)/\ker(\psi) = (G/N)/(K/N) \cong G/K$$ Assertion (5.) then follows.

Community
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Akira
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1 Answers1

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Your proof(s) seem fine to me.

However, you have used "clearly" three times. Whenever you feel the urge to use such a word in mathematics, imagine a complete beginner asking, "why?" - because these are the places we most often hide gaps in our own understanding. Don't fool yourself!

Shaun
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