Jtdylktuy

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

$(1) forall g in G, hin H$ 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

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) $forall g in G$, $hin H$, 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 $hin H$. Then $ghg^{-1}in H$, so $ghin Hg$; therefore $gHsubseteq Hg$. Also $g^{-1}hgin H$, and therefore $hgin gH$; therefore $Hgsubseteq Hg$.

(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 $gin G$ and $hin H$, then $gh=h'g$, for some $h'in H$, so $ghg^{-1}=h'gg^{-1}=h'in H$.

(5)$implies$(1) Let $varphicolon Gto G'$ be a homomorphism with kernel $H$; if $gin G$ and $hin H$, then $varphi(ghg^{-1})=varphi(g)varphi(h)varphi(g^{-1})=varphi(gg^{-1})=1$, so $ghg^{-1}inkervarphi=H$.

(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 $xmapsto xH$ 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 $xyHsubseteq x_1y_1H$. Let $hin H$. Then, denoting by $h_1$ and $h_2$ suitable elements of $H$, we have
$$
xyh=xy_1h_1=xunderbrace{(y_1h_1y_1^{-1})}_{in H}y_1=x_1h_2y_1
=x_1y_1underbrace{(y_1^{-1}h_2y_1)}_{in H}in x_1y_1H
$$