Theorem: Let $G$ be a topological space, and $\mu : G \times G \rightarrow G$ a continuous map. Let $e \in G$ be such that:

• For any $a \in G$, $\mu(a,e) = \mu(e, a) = a$
• $$i^x(a) = (a,e)$$ and $$i^y(a) = (e,a)$$ for any $$a \in G$$

Where $i^x$ and $i^y$ are defined as in 7.4. Then, for any $\beta,\gamma \in \pi(G,e)$, $\mu_* ( i^x_* \beta \cdot i^y_* \gamma) = \beta \cdot \gamma$.

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Proof: Because $\mu$ is continuous, we have $\mu_* (i^x_* \beta \cdot i^y_* \gamma) = \mu_*(i^x_* \beta) \cdot \mu_*(i^y_* \gamma)$. Now, consider that $(i^x_* \beta)(t) = (e, \beta(t))$, by the second property. Then, $(\mu_* i^x_* \beta)(t) = \beta(t)$, by the first property. That is, $\mu_* i^x_* \beta = \beta$, and a symmetric argument obtains $\mu_* i^y_* \gamma = \gamma$. Then, $\mu_* (i^x_* \beta \cdot i^y_* \gamma) = \beta \cdot \gamma$.

By the last exercise (7.4), $i^x_* \beta \cdot i^y_* \gamma$ and $i^y_* \gamma \cdot i^x_* \beta$ commute, so for any $\beta, \gamma \in \pi(G,e)$, we have

$$\beta \cdot \gamma = \mu_*(i^x_* \beta \cdot i^y_* \gamma) = \mu_*(i^y_* \gamma \cdot i^x_* \beta) = \gamma \cdot \beta$$

So, $\pi(G,e)$ is an abelian group.