If G/Z(G) is cyclic, then G is abelian

The Mathematician — Sat, 29 Oct 2022 13:00:46 +0000

If $G/Z(G)$ is cyclic, then $G$ is abelian.

Proof. Recall that that center of

G

is defined as

\begin{align*} Z(G) = \{z \in G \ | \ \forall g \in G, gz = zg\}. \end{align*}

We have that

G/Z(G)

is cyclic, so there exists an

x \in G

such that

\begin{align*} G/Z(G) = \langle xZ(G) \rangle \end{align*}

where

x

is the representative of the coset

xZ(G)

. Let

g \in G

. Then

gZ(G) = (xZ(G))^m = x^mZ(G)

for some

m \in \mathbb{Z}

, where the second equality is proven here because

Z(G)

is a normal subgroup of

G

. Further, we know that

aH = bH

if and only if

b^{-1}a \in H

, where

a,b \in G

and

H

is a subgroup of

G

. So since

gZ(G) = x^mZ(G)

, we have that

(x^m)^{-1}g \in Z(G)

. Therefore, there exists an

z \in Z(G)

such that

(x^m)^{-1}g = z

if and only if

g = zx^m

. After seeing the above, we can finally prove that

G

is abelian. Take

g_1,g_2 \in G

such that

g_1 = x^kz_1

and

g_2 = x^lz_2

where

k,l \in \mathbb{Z}

and

z_1,z_2 \in Z(G)

. So we get

\begin{align*} g_1g_2 &= x^kz_1x^lz_2 \\ &= x^kx^lz_1z_2 \quad \text{because } z_1x^l = x^lz_1 \text{ as } z_1 \in Z(G)\\ &= x^{k + l}z_1z_2 \\ &= x^lx^kz_1z_2 \\ &= x^lx^kz_2z_1 \quad \text{because } z_1z_2 = z_2z_1 \text{ as } z_1,z_2 \in Z(G) \\ &= x^lz_2x^kz_1 \quad \text{because } x^kz_2 = z_2x^k \text{ as } z_2 \in Z(G) \\ &= g_2g_1, \end{align*}

where we have proven that

G

is abelian.

The post If G/Z(G) is cyclic, then G is abelian appeared first on Epsilonify.

If g^2 = e for all g in G, then G is abelian

The Mathematician — Sun, 23 Oct 2022 13:00:41 +0000

If $g^2 = e$ for all $g$ in $G$ , then $G$ is abelian

Proof. Assume that for all

g

G

that

g^2 = e

holds. Let

a,b \in G

. Then since

G

is a group, we know that

ab \in G

. Further, we know that

a^2 = e

and that

b^2 = e

. Now take

(ab)^2 = abab

. Because

(ab)^2 = e

, we have that

abab = e

. Now, multiply

b^{-1}a^{-1}

on both sides on the right side of

abab = e

, then we get

ab = b^{-1}a^{-1}

. As

a^2 = e

, we have that

a = a^{-1}

. The same holds argument for

b

. So we get

ab = b^{-1}a^{-1} = ba

. Wrapping everything together, we can write the proof as

\begin{align*} (ab)^2 = e &\iff abab = e \\ &\iff ababb^{-1} = b^{-1} \\ &\iff aba = b^{-1} \\ &\iff abaa^{-1} = b^{-1}a^{-1} \\ &\iff ab = b^{-1}a^{-1} \\ &\iff ab = ba \quad \text{because } a = a^{-1} \text{ and } b = b^{-1}. \end{align*}

The post If g^2 = e for all g in G, then G is abelian appeared first on Epsilonify.

then G is abelian Archives - Epsilonify

If G/Z(G) is cyclic, then G is abelian

If g^2 = e for all g in G, then G is abelian