Are G-modules the same as left modules over the corresponding group ring?

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For context, I came across the notion of a $G$-module recently when reading about group actions and I'm trying to map it back to concepts from ring theory which are more familiar to me.

A $G$-module is defined on Wikipedia as an Abelian group $M$, together with a group $G$ so that $G$ acts compatibly on $M$.

  1. $M$ is an Abelian group.
  2. $g \vartriangleright (a+b)$ is $(g \vartriangleright a) + (g \vartriangleright b)$.
  3. $gh \vartriangleright a$ is $g \vartriangleright (h \vartriangleright a)$.

It seems to me, however, that we wouldn't lose anything by considering left $\mathbb{Z}[G]$-modules instead. Let $r, s$ be understood to refer to elements of $R=\mathbb{Z}[G]$.

  1. $M$ is an Abelian group.
  2. $r(a+b)$ is $ra + rb$.
  3. $(rs)a$ is $r(sa)$.
  4. $1a$ is $a$.

An element of $\mathbb{Z}[G]$ has the form:

  • $g_1 + g_2 + g_3 + \cdots + g_n$

And thus: $(g_1 + g_2 + g_3 + \cdots + g_n)a = (g_1 \vartriangleright a) + (g_2 \vartriangleright a) + (g_3 \vartriangleright a) + \cdots + (g_n \vartriangleright a)$

And likewise:

  • $g_1a = g_1 \vartriangleright a$