Question

Let ${\mathbf{G}}_{\mathbf{1}}\mathbf{,}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{,}{\mathbf{G}}_{\mathbf{n}}$be groups. Prove that ${\mathbf{G}}_{\mathbf{1}}\mathbf{\times }\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{\times }{\mathbf{G}}_{\mathbf{n}}$is abelian if and only if every ${\mathbf{G}}_{\mathbf{i}}$is abelian.

Short answer

Answer:

It is proved that, $G_1\times ...\times G_n$is abelian if and only if every $G_i$is abelian.

Step-by-step solution

Condition for abelian and non-abelian group

The group $\mathbf{G}$is said to be abelian if $\mathbf{ab}\mathbf{=}\mathbf{ba}$for all $\mathit{a}\mathit{b}\mathbf{\in }\mathit{G}$, and the group $\mathit{G}$is said to be non-abelian if $\mathit{a}\mathit{b}\mathbf{\ne }\mathit{b}\mathit{a}$for all $\mathit{a}\mathbf{,}\mathit{b}\mathbf{\in }\mathit{G}$.

Prove that every Gi is abelian

First show that if $G_1\times ...\times G_n$is abelian then each $G_i$is abelian.

Consider $\left(a_1,a_2,...,a_n\right),\left(b_1,b_2,...,b_n\right)\in G_1\times G_2\times ...\times G_n$, where each $a_i,b_i\in G_i$.

Let $G_1\times ...\times G_n$is abelian.

Then, role="math" localid="1656467998409" $\left(a_1,a_2,...,a_n\right).\left(b_1,b_2,...,b_n\right)=\left(b_1,b_2,...,b_n\right).\left(a_1,a_2,...,a_n\right)$.

Therefore, role="math" localid="1656467974311" $\left(a_1.b_1,a_2.b_2,...,a_n.b_n\right)=\left(b_1.a_1,b_2.a_2,...,b_n.a_n\right)$.

This implies,

$$\begin{gathered} a_1.b_1=b_1.a_1 \\ {a}_2.b_2=b_2.a_2 \\ ⋮ \\ {a}_n.b_n=b_n.a_n \end{gathered}$$

Hence, $a_i.b_i=b_i.a_i$for each $a_i{b}_i\in G_i$.

Since $a_i{b}_i\in G_i$were arbitrary chosen, this implies each $G_i$is abelian.

Prove that G1×...×Gn is abelian

Now, show that if every $G_i$is abelian then $G_1\times ...\times G_n$is abelian.

Consider$a_i{b}_i\in G_i$.

Every $G_i$is abelian this implies that $a_i.b_i=b_i$$.a_i$for all $a_i{b}_i\in G_i$.

Consider $\left(a_i,a_2,...,a_n\right).\left(b_1,b_1,...,b_n\right)=\left(a_1.b_1,a_2.b_2,...,a_n.b_n\right)$.

Then,

$$\begin{gathered} a_1.b_1=b_1.a_1 \\ {a}_2.b_2=b_2.a_2 \\ ⋮ \\ {a}_n.b_n=b_n.a_n \end{gathered}$$

This implies as:

$$\begin{gathered} \left(a_1.b_1,a_2.b_2,...,a_n.b_n\right)=\left(b_1.a_1,b_2.a_2,...,b_n.a_n\right) \\ =\left(b_1,b_2,...,b_n\right).\left(a_1,a_2,...,a_n\right) \end{gathered}$$

Thus, $\left(a_1,a_2,...,a_n\right).\left(b_1,b_2,...,b_n\right)=\left(b_1,b_2,..,b_n\right).\left(a_1,a_2,...,a_n\right)$.

Since,$\left(a_1,a_2,...,a_n\right),\left(b_1,b_2,...,b_n\right)\in G_1\times G_2\times ...\times G_n$were arbitrarily chosen.

This implies that, $G_1\times G_2\times ...\times G_n$is abelian.

Therefore, it is proved that $G_1\times ...\times G_n$is abelian if and only if every $G_i$is abelian.