MHB Does (ab)^n Always Equal a^n*b^n in an Abelian Group?

[evinda]

Hello! (Wave)

Let $G$ a multiplicative abelian group. Show with induction that for each positive integer $n$, $(a \cdot b)^n=a^n \cdot b^n, \forall a,b \in G$. Then,show that the same relation stands for $n \leq 0$.

That's what I have tried:

For $n=1$: $(a \cdot b)^1=a^1 \cdot b^1 \checkmark$

Suppose that the relation stands for $n=k$, $(a \cdot b)^k=a^k \cdot b^k$

We want to show that the relation stand for $n=k+1$.
$$(a \cdot b)^{k+1}=a^{k+1} \cdot b^{k+1} \Rightarrow (a \cdot b)^{k} \cdot (a \cdot b)=a^k \cdot a \cdot b^k \cdot b \Rightarrow a^k \cdot b^k \cdot a \cdot b= a^k \cdot a \cdot b^k \cdot b \Rightarrow b^k \cdot a= a \cdot b^k \Rightarrow a \cdot b^k=a \cdot b^k \checkmark $$

Now we have shown that $(a \cdot b)^n=a^n \cdot b^n, n \in \mathbb{Z}^{+} $

For $n \leq 0$: $n=-k, k\in \mathbb{Z}^{+}$
$$(a \cdot b)^n=a^n \cdot b^n \Rightarrow (a \cdot b)^{-k}=a^{-k} \cdot b^{-k} \Rightarrow ((a \cdot b)^{-1})^{k}=(a^{-1} \cdot b^{-1})^{k} \Rightarrow (b^{-1} \cdot a^{-1})^{k}=(a^{-1} \cdot b^{-1})^{k} \Rightarrow (b^{-1})^k \cdot (a^{-1})^k= (a^{-1})^k \cdot (b^{-1})^k \Rightarrow (a^{-1})^k \cdot (b^{-1})^k=(a^{-1})^k \cdot (b^{-1})^k \checkmark$$

For $n=0$: $(a \cdot b)^0=a^0 \cdot b^0 \Rightarrow 1=1 \checkmark$

Could you tell me if it is right??

 

[Ackbach]

Hello! (Wave)

Let $G$ a multiplicative abelian group. Show with induction that for each positive integer $n$, $(a \cdot b)^n=a^n \cdot b^n, \forall a,b \in G$. Then,show that the same relation stands for $n \leq 0$.

That's what I have tried:

For $n=1$: $(a \cdot b)^1=a^1 \cdot b^1 \checkmark$

Suppose that the relation stands for $n=k$, $(a \cdot b)^k=a^k \cdot b^k$

We want to show that the relation stand for $n=k+1$.
$$(a \cdot b)^{k+1}=a^{k+1} \cdot b^{k+1} \Rightarrow (a \cdot b)^{k} \cdot (a \cdot b)=a^k \cdot a \cdot b^k \cdot b \Rightarrow a^k \cdot b^k \cdot a \cdot b= a^k \cdot a \cdot b^k \cdot b \Rightarrow b^k \cdot a= a \cdot b^k \Rightarrow a \cdot b^k=a \cdot b^k \checkmark $$

This is confusing. Your first statement, $(ab)^{k+1}=a^{k+1}b^{k+1}$, is what you're trying to prove! You can't start with that. Perhaps you should reverse the direction of your arrows? I think you could - your steps seem reversible to me.

Now we have shown that $(a \cdot b)^n=a^n \cdot b^n, n \in \mathbb{Z}^{+} $

For $n \leq 0$: $n=-k, k\in \mathbb{Z}^{+}$
$$(a \cdot b)^n=a^n \cdot b^n \Rightarrow (a \cdot b)^{-k}=a^{-k} \cdot b^{-k} \Rightarrow ((a \cdot b)^{-1})^{k}=(a^{-1} \cdot b^{-1})^{k} \Rightarrow (b^{-1} \cdot a^{-1})^{k}=(a^{-1} \cdot b^{-1})^{k} \Rightarrow (b^{-1})^k \cdot (a^{-1})^k= (a^{-1})^k \cdot (b^{-1})^k \Rightarrow (a^{-1})^k \cdot (b^{-1})^k=(a^{-1})^k \cdot (b^{-1})^k \checkmark$$

For $n=0$: $(a \cdot b)^0=a^0 \cdot b^0 \Rightarrow 1=1 \checkmark$

Could you tell me if it is right??

The way you're writing this is not easy to follow. Put one step on one line. Or just do one long equality.

We have that $(ab)^{-1}=b^{-1}a^{-1}=a^{-1}b^{-1}$, giving the base case. Assume the result is true for $-k$, where $k\in\mathbb{Z}^{+}$. That is, assume $(ab)^{-k}=a^{-k}b^{-k}$. Then
\begin{align*}
(ab)^{-(k+1)}&=((ab)^{k+1})^{-1} \\
&=((ab)^k(ab))^{-1} \\
&=(ab)^{-1}(a^kb^k)^{-1} \\
&=b^{-1}a^{-1}(b^k)^{-1}(a^k)^{-1} \\
&=a^{-1}a^{-k}b^{-1}b^{-k} \\
&=a^{-(k+1)}b^{-(k+1)},
\end{align*}
as required.