Let $M$ and $G$ be groups. We call $M$ a $G$-group (or group with operators) if every $g \in G$ corresponds to an endomorphism of $M$, i.e. we have $$ (mn)^g = (n^g)(m^g). $$ (the application of elements from $G$ in exponential notation). Now let $M$ be abelian, and we write it additively. If $R$ is an associative and on both sides distributive ring with unit $1$, and $M$ is an $R$-group in the above sense, i.e we have

(1) $(m_1 + m_2)r = m_1 r + m_2 r$

for all $r \in R$ and $m_1, m_2 \in M$. And suppose further we have

(2) $m1 = m$

(3) $(mr)r' = m(rr')$

(4) $m(r+r') = mr + mr'$

for $m \in M$, $r,r' \in R$. Then we call $M$ an $R$-right module. I have a question on the following statement:

If $M$ is an abelian $G$-group with $m1 = m$ for all $m \in M$, then $M$ is a $\mathbb Z[G]$-module by $$ m \sum_{g \in G} a_g g := \sum_{g\in G} a_g (mg). $$

I do not understand how $M$ becomes a $\mathbb Z[G]$-module, if we just have $(m_1 + m_2)r = m_1 r + m_2r$ and $m1 = m$. For example why (3) of the above definition holds, I do not even see that for $g, h \in G$ we have $(mg)h = m(gh)$?

These definitions are taken almost verbally from B. Huppert, Endliche Gruppen, unfortunately the relevant pages are not visible from google books. My other recent question about g linear actions was related to thinkings about the above statement. And of course I see that every abelian group could be regarded as a $\mathbb Z$-module (in a unique way, as if $M$ is a $\mathbb Z$-module, then $m1 = m$, hence $m(k+1) = mk + m$ and $m - m = 0 = (1-1)m = m - (-1)m$, hence $-m = (-1)m$, which gives that $mk = m + \ldots + m$ ($k$-times), i.e. its equals the induced $\mathbb Z$-module structure). So abelian groups and $\mathbb Z$-modules are in essence the same. But I do not see if this might be helpful here.


With the definitions you give, the statement you are asking about can very very easily be false. As you observe, it will be false for any $G$-group structure on $M$ such that $(mg)h\neq m(gh)$ for some $m\in M$ and $g,h\in G$. It is trivial to come up with an example of such a $G$-group; for instance, you could take $G$ to be any nontrivial group, $M=\mathbb{Z}$, and define $m1=m$ and $mg=2m$ for each $m\in M$ and each $g\in G\setminus\{1\}$. Then if $g,h\in G\setminus\{1\}$, $(mg)h=4m$ for all $m$, but $m(gh)$ is either $2m$ or $m$ (depending on whether $gh=1$).

That said, I am 99% certain that the intended definition of "$G$-group" includes the condition that $(mg)h=m(gh)$. While this is not the part of the general definition of a group with a set of operators, when the collection of operators forms a group, you want to add this condition as an additional axiom in the definition (otherwise, why are you bothering to consider the set of operators as a group at all; you're not using its group structure in any way). If Huppert's book didn't state this in its definition, this was an error.

  • $\begingroup$ Thanks for your answer! But $m1 = m$ was assumed, but not $(mg)h = m(gh)$, so your example is not valid. But what might work is $G = \{1,g\}$ with $mg = 2m$, then $m(gg) = m(1) = m$, but $(mg)g = 4m$. $\endgroup$ – StefanH Jan 24 '16 at 11:01
  • $\begingroup$ Sure, that works. I've edited my example to not use the identity of $G$. $\endgroup$ – Eric Wofsey Jan 24 '16 at 11:05
  • $\begingroup$ Okay, thanks. I thought about the reverse, i.e. requiring $(mg)h = m(gh)$ but dropping $m1 = m$, but could not find any example, I opened a new question for it: math.stackexchange.com/questions/1624791/… $\endgroup$ – StefanH Jan 24 '16 at 11:25

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