Singular chain complex as a graded algebra

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Let $X$ be a topological space and denote $S_*(X)$ the singular chain complex of $X$.

There is a chain map (The Eilenberg-Zilber map)

$$E: S_*(X)\otimes S_*(X) \rightarrow S_*(X \times X)$$

which satisfies nice properties. (See, for example these notes)

Now suppose $X= G$ is a topological group, and denote by $m: G\times G \rightarrow G$ the group multiplication. This map induces a map at the chain level $S_*(m): S_*(G\times G) \rightarrow S_*(G)$.

Therefore, composing the two above maps we get a product structure on $S_*(G)$; namely

$\cdot: S_*(G)\otimes S_*(G) \rightarrow S_*(G)$

that makes $(S_*(G).\cdot)$ into a DGA.

Now I am trying to look at an specific example. Let $G=\{e,\tau\} $ be a finite group of order $2$.

So the chain complex $S_*(X)$ is generated by the only two 0-simplices $\sigma_e, \sigma_\tau$. By the properties of the map $E$, we get

  • $E(\sigma_\tau, \sigma_\tau) = \sigma_{(\tau,\tau)}$

where the latter denotes the 0-simplex in $G \times G$. Then composing with the map $m$ we get the 0-simplex $\sigma_e$ in $G$. Therefore, $\sigma_\tau^2 = \sigma_e$.

I just want to make sure that my example is correct so far, and I am trying to figure out an algebraic description of the DGA $(S_*(G), \cdot)$ maybe as some sort of exterior algebra.