In an exam I needed to prove that the set of all the endomorphisms of $\mathbb{R}^3$ form an abelian group (I take endomorphisms as linear transformations from a Vector Space to itself).

I tried to answer using the usual sum of functions and seemed correct, but I believe it was too easy that way. To try to make that set a group with the usual function composition, how can I guarantee the existence of inverses? More than that, how can I prove conmutativity?

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    $\begingroup$ $R=\operatorname{End}(\Bbb R^3)$ is a ring, i.e., its additive group $(R,+)$ is abelian. $\endgroup$ – Dietrich Burde Oct 16 '19 at 18:05
  • $\begingroup$ Strictly speaking, the problem statement should have specified the group operation (i.e., pointwise addition of maps) unless it is unambiguous what the operation has to be. In the presence of several natural compositions (here: also composition) this is certainly required for clarity $\endgroup$ – Hagen von Eitzen Oct 16 '19 at 18:11

The set of endomorphisms for $(\mathbb{R}^3,+)$ will form an abelian group under the operation of (pointwise) addition of functions and not with composition, i.e. $$(f+g)(x)=f(x)+g(x).$$ In which case the inverse will be $-f(x)$ and commutativity is the outcome of commutative addition in $\mathbb{R}^3$.

In general, if we have $(G,+)$ an abelian group then the set of endomorphisms of $G$ will form an abelian group under function addition.

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