# Equivalent definitions of normal subgroups

Those 5 Statements are all equivalent to the Statement that $$H$$ is a normal subgroup of $$G$$:

$$(1) \forall g \in G, h\in H$$ we have $$ghg^{-1} \in H$$

$$(2) \forall g \in G, gHg^{-1} \subseteq H$$

$$(3) \forall g \in G, gH = Hg$$

$$(4)$$ Every right coset of $$H$$ is a left coset

$$(5) H$$ is the kernel of a homomorphism of $$G$$ to some other Group

In class we used the Definition 3 can someone explain me why 3 and 1 are equivalent?

Edit: It would be convenient to show the equivalence of all 5 statements

• Try to prove it on your own... It is not that hard! – TheGeekGreek Dec 17 '18 at 13:34
• Welcome to Math.SE! Did you try searching this online? Please see if ProofWiki helps. – GNUSupporter 8964民主女神 地下教會 Dec 17 '18 at 13:36
• May be you can see the prove in book "Topics in Algebra" (I. N. Herstein). In that book, (1) is definition and (3) is lemma. If you have 2nd edition then you can see in page 50 and page 51. – AfroditDione142 Dec 17 '18 at 20:13
• @RM777: Check out "A Course in Group Theory" by J. F. Humphreys (ISBN 0198534590). It is basic and comprehensive. – Moritz Dec 18 '18 at 18:33
• (Also, engage with the existing answer. Its rude to simply ignore it and ask for more answers. Certainly I would be unwilling to put in any effort here if you are just going to ignore the existing, fine answer!) – user1729 Feb 5 at 12:24

Assume (4).

We first show

(4) $$\Longrightarrow$$ (3).

Proof. Let $$g \in G$$. Since every right coset of $$H$$ is a left coset, there exists a $$b \in G$$ such that $$Hg=bH$$. Clearly, it is $$g \in Hg=bH$$ and so $$g^{-1}b \in H$$. From this it follows

$$Hg=(gg^{-1})(bH)=g((g^{-1}b)H)=gH$$

which shows (3).

So from now, we use the fact that $$gH=Hg$$ for every $$g \in G$$. Now we proof (4) $$\Longrightarrow$$ (5). First we show

If $$aH \neq bH$$, then $$aH \cap bH=\emptyset$$.

Proof. We show: If $$aH \cap bH \neq \emptyset$$, then $$aH=bH$$. Especially, it then follows

$$b \in aH \Longrightarrow aH=bH.$$

So let $$g \in aH \cap bH$$, i. e. $$g=ak$$ with $$k \in H$$ and $$g=bl$$ with $$l \in H$$. Then we have for every $$h \in H$$ that $$ah=blk^{-1}h \in bH$$, so $$aH \subseteq bH$$. In the same way it follows $$bH \subseteq aH$$ and so the claim.

The proof that "$$\sim$$" is an equivalence relation you can read here.

Next we show

$$G$$ is the disjoint union of its left cosets (right cosets).

Proof. Since every $$a \in G$$ is an element of the left coset $$aH$$ (right coset $$Ha$$), $$G$$ is the union of its cosets. Since our preceding result, this union is disjoint.

We are now ready to state our final result, which proofs (5). In the following, $$G/H=\{aH: a \in G\}$$ denotes the set of left cosets of $$H$$ and ker denotes the kernel of a homomorphism.

If $$G$$ is a group and $$H$$ is a normal subgroup of $$G$$, then $$G/H$$ is a group with respect to the multiplication $$aH \cdot bH :=abH$$ and $$\varphi: G \longrightarrow G/H, \ a \longmapsto aH$$ is a surjective group homomorphism such that $$\operatorname{ker}(\varphi)=H$$.

Proof. At first we have to show that the multiplication if well-defined, i. e. $$abH=a'b'H$$ for all $$a,a',b,b' \in G$$ such that $$aH=a'H$$ and $$bH=b'H$$. Let $$a'=ah$$ and $$b'=bk$$ with $$h,k \in H$$. Since $$H$$ is a normal subgroup of $$G$$, there exists an $$h' \in H$$ such that $$hb=bh'$$. It follows

$$a'b'H=ahbkH=abh'hH=abH.$$

Now we show that "$$\cdot$$" is associative. This is very clear, since

$$(aH \cdot bH)\cdot cH=abH \cdot cH=abcH=aH \cdot (bcH)=aH \cdot (bH \cdot cH)$$

holds for every $$aH, bH, cH \in G/H$$. It is $$eH=H \in G/H$$ the neutral element because

$$eH \cdot aH=eaH=aH=aeH= aH \cdot eH$$

works vor every $$aH \in G/H$$. The inverse of an $$aH \in G/H$$ is $$a^{-1}H \in G/H$$ since we have

$$aH \cdot a^{-1}H=aa^{-1}H=eH=H=eH=a^{-1}aH=a^{-1}H \cdot aH$$

for every $$aH \in G/H$$. So we have shown that $$G/H$$ is a group. Next we show that $$\varphi$$ is a group homomorphism. For every $$a,b \in G$$ we have

$$\varphi(ab)=abH=aH \cdot bH=\varphi(a)\varphi(b).$$

Because of $$\varphi(a)=aH$$, $$\varphi$$ is clearly surjective. Finally we have to show that $$\operatorname{ker}(\varphi)=H$$. This follows from

$$a \in \operatorname{ker}(\varphi) \Longleftrightarrow \varphi(a)=eH \Longleftrightarrow aH=eH \Longleftrightarrow a \in H.$$

Remark. $$G/H$$ is called the quotient group of $$G$$ by $$H$$.

Now it remains to show (5) $$\Longrightarrow$$ (1).

Let $$G,G'$$ be groups and $$\varphi: G \longrightarrow G'$$ be a group homomorphism. Define $$H:=\operatorname{ker}(\varphi)$$. Then $$ghg^{-1} \in H$$ for all $$g \in G$$ and $$h \in H$$.

Proof. Let $$gh \in gH$$ with $$h \in H$$. Then it follows

$$\varphi(ghg^{-1})=\varphi(g)\varphi(h)\varphi(g^{-1})=\varphi(g)\varphi(g^{-1})=e_{G'},$$

so $$ghg^{-1} \in H$$.

Assume $$(1)$$ and let $$g \in G$$.

Take an arbitrary $$gh \in gH$$ for $$h \in H$$. Using $$(1)$$ we get $$ghg^{-1} \in H$$ so $$\exists h' \in H$$ such that $$ghg^{-1} = h'$$. This means $$gh = h'g \in Hg$$. Since $$gh \in gH$$ was arbitrary, we conclude $$gH \subseteq Hg$$.

To show the converse inclusion, take an arbitrary $$hg \in Hg$$ for $$h \in H$$. Then $$g^{-1}h \in g^{-1}H \subseteq Hg^{-1}$$ so $$\exists h' \in H$$ such that $$g^{-1}h = h'g^{-1}$$. Mulyiplying this with $$g$$ from both sides yields $$hg = gh' \in gH$$. Therefore $$Hg \subseteq gH$$.

Since $$g \in G$$ was arbitrary, $$(3)$$ follows.

Assume $$(3)$$ and let $$g \in G, h \in H$$ be arbitrary.

Since $$(3)$$ holds, we have $$gh \in gH = Hg$$ so $$\exists h' \in H$$ such that $$gh = h'g$$. Multiplying by $$g^{-1}$$ from the right yields $$ghg^{-1} \in H$$. Thus $$(1)$$ holds.

Theorem. Let $$G$$ be a group and $$H$$ a subgroup of $$G$$. The following conditions are equivalent

(1) $$\forall g \in G$$, $$h\in H$$, we have $$ghg^{-1} \in H$$

(2) $$\forall g \in G$$, $$gHg^{-1} \subseteq H$$

(3) $$\forall g \in G, gH = Hg$$

(4) Every right coset of $$H$$ is a left coset

(5) $$H$$ is the kernel of a homomorphism of $$G$$ to some other group

Proof. (1)$$\implies$$(2) Obvious.

(2)$$\implies$$(3) Let $$h\in H$$. Then $$ghg^{-1}\in H$$, so $$gh\in Hg$$; therefore $$gH\subseteq Hg$$. Also $$g^{-1}hg\in H$$, and therefore $$hg\in gH$$; therefore $$Hg\subseteq Hg$$.

(3)$$\implies$$(4) Obvious.

(4)$$\implies$$(1) Since distinct left cosets are disjoint, if $$gH=Hg'$$, then both sets contain $$g$$; therefore $$Hg'=Hg$$. If $$g\in G$$ and $$h\in H$$, then $$gh=h'g$$, for some $$h'\in H$$, so $$ghg^{-1}=h'gg^{-1}=h'\in H$$.

(5)$$\implies$$(1) Let $$\varphi\colon G\to G'$$ be a homomorphism with kernel $$H$$; if $$g\in G$$ and $$h\in H$$, then $$\varphi(ghg^{-1})=\varphi(g)\varphi(h)\varphi(g^{-1})=\varphi(gg^{-1})=1$$, so $$ghg^{-1}\in\ker\varphi=H$$.

(1)$$\implies$$(5) Once we prove that the operation $$(xH)(yH)=(xy)H$$ on the set of left cosets is well defined, then it obviously defines a group structure, with $$x\mapsto xH$$ being a homomorphism with kernel $$H$$.

Let $$xH=x_1H$$ and $$yH=y_1H$$; we want to show that $$xyH=x_1y_1H$$. By symmetry, we just need to show that $$xyH\subseteq x_1y_1H$$. Let $$h\in H$$. Then, denoting by $$h_1$$ and $$h_2$$ suitable elements of $$H$$, we have $$xyh=xy_1h_1=x\underbrace{(y_1h_1y_1^{-1})}_{\in H}y_1=x_1h_2y_1 =x_1y_1\underbrace{(y_1^{-1}h_2y_1)}_{\in H}\in x_1y_1H$$