An equivalence relation on the isomorphism classes of finite groups Let all groups be finite. I am interested in the equivalence relation on finite groups given by $G \sim H$, if and only if, there exist $W_1,W_2,Z$ with $W_1 \cong W_2 \unlhd Z$ so that $G \oplus Z/W_1 \cong H\oplus Z/W_2$.
The most obvious question to ask is when are two groups equivalent. A necessary condition is that they have the same order. In the case, if the two groups are abelian this is sufficient. This can be worked out from the classification of finite abelian groups and the fact $\mathbb{Z}_{p^n} \sim \mathbb{Z}_{p^n-1} \oplus \mathbb{Z}_p$. This means abelian groups under this relation are isomorphic to the positive integers under multiplication.
Can anyone provide two groups of the same order that are not equivalent? I was not able to do it for the case $S_3,\mathbb{Z}_6$.
The reason I am interested in the question is because I recently saw the definition of topological K-theory. I was interested if there was some interesting way to mimic it for the case of groups. For various reasons, I decided that before forming the Grothendieck group, I wanted the elements to be of the form $G-H$ (with addition component wise direct sum) where $H \leq G$. It should also have the property that $G+H-H=G$. The idea behind this is that for any subgroup $H$ we are formally introducing another subgroup to act as a complementary summand to $H$. However, if the subgroup already is a direct summand these should coincide. But if we are treating subgroups like direct summands, it only makes sense to also require that if $H \unlhd G$, then $G-H \sim G/H$, from which you get things like $\mathbb{Z}_{p^n} \sim \mathbb{Z}_{p^n-1} \oplus \mathbb{Z}_p$.
It was a hassle to find the equivalence relation generated by this, but it ended up being: $G_1 - H_1 \sim G_2 - H_2$, if and only if, there exists $H_1,H_2 \unlhd F$ and $W\cong\bar{W} \unlhd Z$ so that $F/H_1 \oplus G_1 \oplus Z/W \cong F/H_2 \oplus G_2 \oplus Z/\bar{W}$.
So the monoid I am interested in is the set of formal differences $G-H$ with $H \leq G$, quotiented out by this equivalence relation. The question in the first paragraphs is what you get when you ask when $G \sim H$ in this monoid. It's worth noting that nontrivial inverses don't exist, and the operation remains cancellative even after quotienting. I think this means the Grothendieck group will contain pretty much the same information as this monoid.
So to reiterate, $\textbf{Can anyone provide two groups of the same order that are not equivalent?}$. I think a good pair to check might be a perfect group and an abelain group. If this question ends up being too easy, feel free to talk about the structure of the monoid.
 A: As I said in my comment, if two groups are equivalent, then they must have the same composition factors. This follows from the Jordan-Hõlder Theorem, because $W_1 \cong W_2$ implies that $Z/W_1$ and $Z/W_2$ have the same composition factors.
The converse is also true. Any two finite groups with the same composition factors are equivalent. (So in particular any two solvable groups of the same order are equivalent.)
To show this, it is sufficient to show that any finite group is equivalent to the direct product of its composition factors. This is clear for simple groups $G$ so suppose $G$ is not simple, and let $N$ be a minimal nontrivial normal subgroup of $G$. By induction, it is sufficient to show that $G \sim G/N \times N$.
Let $H = G/N \times N$, and let $\sigma:G \to G/N$ and $\tau:H \to G/N$ be the natural homomorphisms with kernel $N$. Let $Z$ be the subgroup $\{(g,h) : \sigma(g) = \tau(h)\}$ of the direct product $G \times H$, and let $W_1 = \{(n,1): n \in N\}$ and $W_2 = \{(1,n):n \in N\}$.
Then $W_1 \cong W_2 \cong N$ with $W/N_1 \cong H$ and $W/N_2 \cong G$, so $G \times W/N_1 \cong H \times W/N_2 \cong G \times H$, and hence $G \sim H$.
