# 10 equivalent definitions of normal subgroup

I've found various equivalent definitions of normal subgroup from this Wikipedia page. I've just finished proving their equivalence. Could you please verify if it is fine or contains logical mistakes?

Let $$N$$ be a subgroup of $$G$$. Then the following statements are equivalent.

a. For all $$g,h \in G$$: $$gh \in N \iff hg \in N$$.

b. For all $$g \in G$$: $$gNg^{-1} \subseteq N$$.

c. For all $$g \in G$$: $$gNg^{-1} = N$$.

d. The sets of left and right cosets of $$N$$ in $$G$$ coincide.

e. For all $$x,y,g,h \in G$$: $$x \in gN$$ and $$y \in hN \implies xy \in (gh)N$$.

f. For all $$n \in N, g \in G$$: $$n^{-1} g^{-1} n g \in N$$.

g. For all $$n \in N, g \in G$$: $$g n g^{-1} \in N$$.

h. $$N = \bigcup_{n \in N} \operatorname{Cl}(n)$$ where $$\operatorname{Cl}(n) := \{gng^{-1} \mid g \in G\}$$.

i. For all $$g \in G$$: $$gN = Ng$$.

j. There exists a group homomorphism whose domain is $$G$$ and kernel is $$N$$.

My attempt: To make it easier to follow, I put each part of the proof between two consecutive definitions.

For all $$g,h \in G$$: $$gh \in N \iff hg \in N$$.

Assume the contrary that there exist $$n\in N$$ and $$g \in G$$ such that $$gng^{-1} \notin N$$. Then $$g^{-1} (gn) = n\notin N$$ by (a). This is a contradiction.

For all $$g \in G$$: $$gNg^{-1} \subseteq N$$.

Substituting $$g$$ for $$g^{-1}$$ in $$gNg^{-1} \subseteq N$$, we get $$g^{-1} N g\subseteq N$$. Substituting $$N$$ for $$g^{-1} N g$$ in $$gNg^{-1} \subseteq N$$ , we get $$N \subseteq g^{-1} N g$$. As a result, $$g^{-1} N g = N$$. Substituting $$g$$ for $$g^{-1}$$ in $$g^{-1} N g = N$$, we get the desired result.

For all $$g \in G$$: $$gNg^{-1} = N$$.

It follows from (c) that $$gN= Ng$$. The result then follows.

The sets of left and right cosets of $$N$$ in $$G$$ coincide.

Because of (d) and the fact that $$h \in hN \cap Nh$$, we have $$hN=Nh$$. It follows from $$x \in gN$$ and $$y \in hN$$ that $$x = gn_1$$ and $$y = hn_2$$ for some $$n_1,n_2 \in N$$. Then $$xy = gn_1 hn_2$$. Because $$Nh = hN$$, $$n_1 h = h n_3$$ for some $$n_3 \in N$$. Then $$xy = g h n_3 n_2 = (gh) (n_3 n_2) \in (gh) N$$.

For all $$x,y,g,h \in G$$: $$x \in gN$$ and $$y \in hN \implies xy \in (gh)N$$.

We have $$n^{-1} g^{-1} n \in n^{-1} g^{-1} N$$ and $$g \in gN$$. Then by (e), we have $$n^{-1} g^{-1} n g \in (n^{-1} g^{-1} g)N = n^{-1} N = N$$.

For all $$n \in N, g \in G$$: $$n^{-1} g^{-1} n g \in N$$.

Substituting $$g$$ for $$g^{-1}$$ in $$n^{-1} g^{-1} n g \in N$$, we get $$n^{-1} g n g^{-1} \in N$$. Because $$n^{-1} \in N$$, we have $$g n g^{-1} \in N$$.

For all $$n \in N, g \in G$$: $$g n g^{-1} \in N$$.

It follows from (g) that $$\operatorname{Cl}(n) \subseteq N$$ for all $$n \in N$$. Then $$\bigcup_{n \in N} \operatorname{Cl}(n) \subseteq N$$. On the other hand, $$n \in \operatorname{Cl}(n)$$ and thus $$N \subseteq \bigcup_{n \in N} \operatorname{Cl}(n)$$. The result the follows.

$$N = \bigcup_{n \in N} \operatorname{Cl}(n)$$ where $$\operatorname{Cl}(n) := \{gng^{-1} \mid g \in G\}$$.

It follows from (h) that $$\operatorname{Cl}(n) \subseteq N$$ for all $$n \in N$$. As a result, for all $$g \in G, n \in N$$, we have $$g n g^{-1} = n'$$ for some $$n' \in N$$. Hence $$gn=n'g$$ for some $$n' \in N$$. Thus $$gN \subseteq Ng$$. By symmetry, we also have $$Ng \subseteq gN$$. The result then follows.

For all $$g \in G$$: $$gN = Ng$$.

Let $$G/N := \{gN \mid g \in G\}$$. We define a binary operation $$G/N \times G/N \to G/N$$ by $$(gN) (hN) \mapsto (gh)N$$. Let's prove that it's well-defined, i.e. $$gN = aN$$ and $$hN = bN$$ implies $$(gh) N = (ab) N$$.

We have $$gN= aN$$ implies $$g=an_1$$ for some $$n_1 \in N$$. Similarly, $$h=bn_2$$ for some $$n_2 \in N$$. Then $$gh = an_1bn_2$$. It follows from (i) that $$Nb=bN$$ and thus $$n_1b=bn_3$$ for some $$n_3 \in N$$. Hence $$gh =abn_3n_2$$. Because $$n_2,n_3 \in N$$, we have $$n_3n_2 \in N$$ and thus $$(n_3n_2)N =N$$. As a result, $$(gh)N = (abn_3n_2)N = (ab)(n_3n_2)N = (ab)N$$.

It's then straightforward to verify that $$G/N$$ together with above operation is group. Now we define a map $$\phi: G \to G/N, g \mapsto gN$$. It's easy to verify that $$\phi$$ is in fact a group homomorphism such that $$\operatorname{ker} \phi = \{g \in G \mid gN = 1N =N\} =N$$.

There exists a group homomorphism whose domain is $$G$$ and kernel is $$N$$.

Let $$\phi: G \to K$$ be such a group homomorphism that $$\operatorname{ker} \phi = N$$. If $$gh \in N$$ then $$\phi (gh) = \phi(g) \phi (h) = 1$$. This means $$\phi(g) = (\phi(h))^{-1}$$. As a result, $$\phi (hg) = \phi(h) \phi (g) = \phi(h) (\phi(h))^{-1} = 1$$ and thus $$hg \in N$$. By symmetry, we have $$hg \in N \implies gh \in N$$. This completes the proof.

For all $$g,h \in G$$: $$gh \in N \iff hg \in N$$.

• In the second one, you can’t “substitute” anything for $N$. You can pick the element you want, but you cannot replace $N$ with something else. Instead, once you have $g^{-1}Ng\subseteq N$, multiply on the left by $g$ and on the right by $g^{-1}$. – Arturo Magidin Jul 7 at 20:46
• @ArturoMagidin Thank you so much for correcting my logical misunderstanding! This is a critical mistake. – LAD Jul 7 at 20:50

In the second one, you can’t “substitute” anything for $$N$$. You can pick the element you want, but you cannot replace $$N$$ with something else. Instead, once you have $$g^{-1}Ng\subseteq N$$, multiply on the left by $$g$$ and on the right by $$g^{-1}$$.