Why does the sum over the group remain unchanged when reindexed by right multiplication with a fixed element?

Conditions

• $G$ is a finite group.
• $h$ is a fixed element in $G$.
• $f$ is any function defined on $G$.

Reasoning, step by step

1. Consider the sum $\sum_{g \in G} f(g)$.
2. Replace the index $g$ with $gh$ for a fixed $h \in G$.
3. Observe that the map $g \mapsto gh$ is a bijection from $G$ to $G$.
4. Conclude that the set of values $\{gh \mid g \in G\}$ is identical to $G$.
5. Deduce that $\sum_{g \in G} f(g) = \sum_{g \in G} f(gh)$ because addition is commutative and the terms are the same.

Example

The video writes $\sum_{g\in G} f(g)=\sum_{g\in G} f(gh)$ and explains that "set of $\{gh\}$ = set of $G$", so the sum is exactly the same.

Common misconceptions

• Thinking that the order of summation matters in a finite sum; it does not.
• Confusing left multiplication with right multiplication; the video uses right multiplication $g \mapsto gh$ but the principle holds for both.

Watch the explanation

Explore next

• How is this reindexing identity used to prove the invariance of the averaged inner product?
• What is the role of the finiteness of $G$ in this argument?