Given an abelian group $G$, one can form the endomorphism ring $\mbox{End}(G)$ by letting $\alpha+\beta=\alpha(x)+\beta(x)$, and $\alpha\beta=\alpha(\beta(x))$, where $\alpha$ and $\beta$ are endomorphisms. Clearly, composition distributes over addition, and addition is commutative, so $\mbox{End}(G)$ is a ring. My question is: when is $\mbox{End}(G)$ commutative? Are there a nice set of criteria, or, if there is no such nice set of criteria, is there a nice class of abelian groups with commutative endomorphism rings.
[Math] Criterion for an abelian group to have a commutative endomorphism ring
gr.group-theoryra.rings-and-algebras
Related Solutions
Here is an approximation of an answer to "For what finite groups is Aut(G) simple?"
As Daniel Miller mentioned, Inn(G) is a normal subgroup of Aut(G), so for Aut(G) to be simple either Inn(G) = 1, in which case G is abelian, or Inn(G) = Aut(G) is simple. The former case should be somewhat easy to handle assuming G is finite. In the latter case, we have that G/Z(G) is simple. If G is also perfect, then G is called quasi-simple. Of course, G need not be perfect as G ≅ A5 × 2 shows. However, I believe this is the only obstruction, so ignoring a possible cyclic direct factor of order 2, G/Z(G) is simple, and G is quasi-simple. The finite quasi-simple groups and their automorphism groups are classified, but the classification is a bit long. For a fixed simple group, X = G/Z(G), there are only finitely many isomorphism classes of quasi-simple groups D such that D/Z(D) = X. In fact there is a unique largest one called the Schur cover, that I'll call D. If Z(D) is cyclic, then in fact Aut(G) = Aut(X) = Aut(D) does not pay any attention to the center. So all we need to do is find all X with Aut(X) = X [and each one works], and all X with Z(D) non-cyclic [and check which ones work].
Having done most, but not all, of that, I thought it might help to record the basic result:
If G = H×T where T=1 if H is abelian and T is cyclic of order dividing 2 otherwise, and where H is on the following list, then Aut(G) is simple:
- cyclic of order 3, 4, or 6
- elementary abelian of order 2n for n ≥ 3
- M11, 2.Sz(8), J1, 2.Sp(6,2), M23, M24, Ru, 2.Ru, Co3, Co2, Ly, Th, Fi23, Co1, 2.Co1, J4, B, 2.B, E7(2), M
- Ω(2n+1,2) for n ≥ 3
- Sp(2n,2) for n ≥ 3
- E8(p) for any prime p
- F4(p) for any prime p
- G2(p) for any prime p ≥ 5
Additionally if Aut(G) is simple, then G = H×T as above, except possibly H/Z(H) is on the following list:
- L3(4), U4(3), U6(2), 2E6(2)
- Ω+(4n,q) for certain q
These are groups with non-cyclic multiplier other than Sz(8) [definitely an example] and Ω+(8,2) [not an example]. The Ω+(4n,q) case should be mostly easy, as there are too many automorphisms to kill. The others would be easy in an ideal world, but as far as I know our computational knowledge of these groups is limited and/or flawed. Of course, I also need to check the abelian case carefully, but I think 3,4,6 and 2^n are the only abelian examples.
It would make another good answer: For what torsion abelian groups G is Aut(G) simple? This would handle the abelian groups here, as well as some of the original poster's interest, without delving into the nastier aspects of abelian groups.
If $G$ is a finite abelian group, then $\mathbb C[G] = \lbrace f \colon \hat G \to \mathbb C \rbrace$, where $\hat G$ is the Pontrjagin dual of $G$. The isomorphism $g \mapsto g^{-1}$ translates into the same map on the Pontrjagin dual (basically multiplication by $-1$ on $\hat G$), but now it is a bit easier to analyze. Note also, that there is a non-canonical isomorphism $G \cong \hat G$.
Hence, in order to find two non-isomorphic abelian groups which have strongly isomorphic complex group rings, we just have to analyze (in addition to the cardinality of the group) the orbit structure of multiplication by $-1$ on the group. Indeed, we are now just talking about an algebra of function on a set with some $\mathbb Z/2 \mathbb Z$-action.
Example: In $\mathbb Z/8\mathbb Z \times \mathbb Z/2\mathbb Z$, there are precisely $4$ elements, which are fixed under multiplication by $-1$, namely $(0,0),(4,0),(0,1)$ and $(4,1)$. The same is true for $\mathbb Z/4\mathbb Z \times \mathbb Z/4\mathbb Z$. Here, one has $(0,0),(2,0),(0,2)$ and $(2,2)$. Hence, there exists an isomorphism between $\mathbb C[\mathbb Z/8\mathbb Z \times \mathbb Z/2\mathbb Z]$ and $\mathbb C[\mathbb Z/4\mathbb Z \times \mathbb Z/4\mathbb Z]$, which respect the isomorphism which is induced by $g \mapsto g^{-1}$.
I do not know about an example with coefficients in $\mathbb Z$.
Best Answer
Amongst the finitely generated abelian groups, those with commutative endomorphism ring are exactly the cyclic groups.
Torsion abelian groups with commutative endomorphism rings are exactly the locally cyclic groups, that is, the subgroups of Q/Z. They were classified in:
This papers also gives more complicated examples of mixed groups with commutative endomorphism ring.
The mixed case was completed in:
This paper indicates the difficulty of any classification of torsion-free abelian groups with commutative endomorphism rings, as Corner has shown that very large torsion-free abelian groups can have commutative endomorphism rings (while the classifications up to now have basically been "only very small ones").