Equivalent definitions of normal subgroups

Question

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

Answer 1

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

Answer 2

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.

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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.

Answer 3

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 $Hg\subseteq gH$.

(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 $$ xyh=xy_1h_1=x\underbrace{(y_1h_1y_1^{-1})}_{\in H}y_1=x_1h_2y_1 =x_1y_1\underbrace{(y_1^{-1}h_2y_1)}_{\in H}\in x_1y_1H $$