Coboundary matrix of bar resolution for group cohomology: do the elementary divisors always divide $|G|$?

Question

Consider the coboundary matrix $C^1(G, \mathbb{Z}) \to C^2(G, \mathbb{Z})$ of the normalized bar resolution of $G$ with coefficients in the trivial $\mathbb{Z}G$-module $\mathbb{Z}$. That is, thinking of the elements of $C^1(G,\mathbb{Z})$ as row vectors in $\mathbb{Z}^{|G|-1}$ (-1 because it's normalized), the coboundary matrix here is of size $(|G|-1) \times (|G|^2-2|G|+1)$.

I noticed that when I put this matrix into Smith Normal Form, it seemed that all the elementary divisors divided $|G|$. I have verified this in GAP for all groups up to and including order 60. (I am happy to share my GAP code and its output if anyone wants. For the curious, this arises in at least one approach to algorithms for testing isomorphism of finite groups.)

Q1. Is it the case that the elementary divisors of the above coboundary map always divide $|G|$, and if so why?

I know that $|G|$ annihilates $H^2(G, \mathbb{Z})$, but that proof is about the transfer map on cohomology and doesn't depend on the resolution, whereas (I think!) the coboundary map above is strongly about this particular resolution, so I couldn't see how the transfer argument would tell me something about this particular coboundary matrix.

---

My second observation is that most of the elementary divisors seem to be 1. (In fact, I originally thought they were always 1 until I did a computer search. At some point I think I had a proof that they are all 1 for cyclic groups.) Up to order 60, most groups have only one or two elementary divisors that aren't 1, and the max number of non-1 elementary divisors is 5 (which occurs only once, for SmallGroup(32, 51)). Also, most elementary divisors seem to not only divide the order of the group, but to be relatively small, even among numbers that divide $|G|$. I know that in the world of finite groups only going up to order 60 one may see patterns due to the "law of small numbers groups", but I nonetheless ask:

Q2. Is there some explanation for the distribution of these elementary divisors, especially why there are so many 1s and why in general they are relatively small among the set of all divisors of $|G|$?

I am particularly curious if for perfect or simple groups, all the elementary divisors are 1 (this is true for $A_5$ and, IIRC, for cyclic groups), but for the sake of not including too many questions in a single post I think I'll leave it at that.

(Note: it is standard that, writing $d(G)$ for the min number of generators of $G$, the columns are all in the $\mathbb{Z}$-span of the columns indexed by pairs $(g,h)$ where $h$ is one of a chosen generating set. Thus we can reduce the size of the matrix to $(|G|-1) \times (|G|-1)d(G)$ without changing the nonzero elementary divisors. This helped a lot in speeding up the GAP calculations.)

Answer 1

Maybe I am missing something, but it seems this is the answer. The elementary divisors of the coboundary map give you the torsion part of the cokernel of the coboundary $\delta$. Now we have an exact sequence $0\to H^n(G,\mathbb Z)\to \mathrm{coker}\ \delta\to \mathrm{im}\delta\to 0$ and image of $\delta$ is free abelian, so this splits as $\mathrm{coker}\ \delta =H^n(G,\mathbb Z)\oplus \mathbb Z^d$ for some $d$, and hence each elementary divisor divides the exponent of $H^n(G,\mathbb Z)$, which as you pointed out divides $|G|$ by the transfer.

Answer 2

Just to complete Benjamin Steinberg's answer for the specific situation in the question, we have $n=2$ and the elementary divisors are those of $H^2(G,\mathbb Z).$ But this group is isomorphic to the abelianization $G/G^\prime$ of $G.$ (This follows from the exact sequence of additive abelian groups $$0\rightarrow\mathbb Z\rightarrow\mathbb Q\rightarrow \mathbb Q/\mathbb Z\rightarrow 0.$$ Since $\mathbb Q$ is divisible, the associated exact sequence in cohomology gives $H^1(G,\mathbb Q/\mathbb Z)\cong H^2(G,\mathbb Z)$ and the first group is the dual group of $G/G^\prime.$)

Alternatively, one can note that if $u_g=1-g$ for $g\in G-1,$ so that $u_g$ forms a basis of the augmentation ideal $I_G\subset\mathbb ZG,$ then the columns of the coboundary matrix in the question give the coefficients of $u_gu_h$ in terms of the $u_k;$ hence the elementary divisors give the quotient $I_G/I_G^2;$ but this is isomorphic to $G/G^\prime$ via $g-1\mapsto gG^\prime.$

Hence in terms of the examples noted in the question, all the elementary divisors will be 1 when $G$ is simple (or more generally, perfect). The example SmallGroup(32,51) is $C_2^5,$ hence again it has five 2's as elementary divisors.