Why Does (a * b)^n = a^n * b^n in an Abelian Group?

In a group, powers are formed by repeatedly applying the group operation. The identity

$$(a * b)^n = a^n * b^n$$

holds in every abelian group because its operation is commutative. The proof for positive exponents is an induction; the zero and negative cases follow from the definitions of group powers and inverses.

What the notation means

Write $e$ for the identity element of $G$. For an element $x \in G$, define $x^0=e$, define $x^n$ for positive $n$ as $n$ repeated copies of $x$, and define $x^{-n}=(x^{-1})^n$ for positive $n$.

The operation is written as $*$ here, so $a*b$ means the group product; it need not be ordinary multiplication. The theorem applies to every integer $n$. If your course uses “natural number” to mean only positive integers, the induction below proves that version directly.

Proof for positive exponents

For $n=1$, the claim is just $a*b=a*b$. Suppose it holds for a positive integer $n$, so that $(a*b)^n=a^n*b^n$. Then

$$\begin{aligned} (a*b)^{n+1} &=(a*b)^n*(a*b) \\ &=a^n*b^n*a*b \\ &=a^n*a*b^n*b \\ &=a^{n+1}*b^{n+1}. \end{aligned}$$

The key reordering is $b^n*a=a*b^n$. Since $G$ is abelian, every pair of elements commutes, including $a$ and $b^n$. This proves the identity for all positive integers by induction.

For $n=0$, both sides are $e$: $(a*b)^0=e$ and $a^0*b^0=e*e=e$.

For $n<0$, write $n=-k$ where $k>0$. In an abelian group, $(a*b)^{-1}=b^{-1}*a^{-1}=a^{-1}*b^{-1}$. Apply the positive-exponent result to $a^{-1}$ and $b^{-1}$:

$$(a*b)^{-k}=((a*b)^{-1})^k=(a^{-1}*b^{-1})^k=(a^{-1})^k*(b^{-1})^k=a^{-k}*b^{-k}.$$

Thus the formula holds for every integer exponent.

Worked example

Consider the nonzero rational numbers under ordinary multiplication. This is an abelian group. Choose $a=2$, $b=3$, and $n=4$. The two sides agree:

$$(2\cdot 3)^4=6^4=1296, \qquad 2^4\cdot 3^4=16\cdot81=1296.$$

The induction proof explains why this works: when the repeated factors are expanded, commutativity lets you collect all four copies of $2$ and all four copies of $3$ separately.

Why commutativity matters

The claim is not true for arbitrary groups. For example, in the permutation group $S_3$, use composition from right to left and let $a=(12)$ and $b=(23)$. Then $a^2=b^2=e$, so $a^2*b^2=e$. But $a*b=(123)$, and $(a*b)^2=(132)\ne e$. Therefore the formula fails for $n=2$ in this nonabelian group.

This also shows why an induction argument cannot simply reorder factors in a general group: $b^n*a$ need not equal $a*b^n$.

Common mistakes

• Assuming the identity holds in every group. It relies on commutativity. Without that condition, the factors in the expanded product may not be regrouped as $a^n*b^n$.
• Ignoring the exponent convention. State whether $n$ is positive or any integer. The theorem is valid for all integers when powers are defined using the identity and inverses.
• Reordering without justification. In the induction step, explicitly use that $a$ commutes with $b^n$; this is guaranteed because the group is abelian.
• Confusing group powers with scalar powers. The notation means repeated use of the group operation. In additive notation, the corresponding statement is $n(a+b)=na+nb$.