Every group is isoclinic to a stem group

Statement
Suppose $$G$$ is a group. Then, there exists a group $$H$$ such that:


 * 1) $$H$$ is a  stem group, i.e., $$Z(H) \le [H,H]$$.
 * 2) $$G$$ and $$H$$ are  isoclinic groups.

Related facts

 * Stem group has the minimum order among all groups isoclinic to it
 * Formula for second cohomology group for trivial group action in terms of Schur multiplier and abelianization

Proof
Given: A group $$G$$.

To prove: There exists a group $$H$$ such that $$G$$ is isoclinic to $$H$$, and $$Z(H) \le [H,H]$$.

Proof: Let $$Z = Z(G)$$ be the center of $$G$$, and let $$W = Z \cap G'$$. First, consider the short exact sequence:

$$0 \to Z \to G \to G/Z \to 1$$

Consider the short exact sequence that gives the Formula for second cohomology group for trivial group action in terms of Schur multiplier and abelianization.

$$0 \to \operatorname{Ext}^1((G/Z)^{\operatorname{ab}};Z) \to H^2(G/Z;Z) \to \operatorname{Hom}(H_2(G/Z;\mathbb{Z}),Z) \to 0$$

The group $$G$$ can be viewed as giving rise to an element of $$H^2(G/Z;Z)$$, which, under the mapping, gives an element of $$\operatorname{Hom}(H_2(G/Z;\mathbb{Z}),Z)$$. It turns out from the constriction (see the formula page) that the image of the mapping must necessarily lie in the subgroup $$Z \cap G'$$. In fact, the image of the homomorphism is precisely $$Z \cap G'$$. Thus, in fact, we get an element of $$\operatorname{Hom}(H_2(G/Z;\mathbb{Z}),Z \cap G')$$.

Now, consider the short exact sequence for the formula for second cohomology group for trivial group action in terms of Schur multiplier and abelianization with $$Z \cap G'$$ used in place of $$Z$$ as the central subgroup:

$$0 \to \operatorname{Ext}^1((G/Z)^{\operatorname{ab}};Z \cap G') \to H^2(G/Z;Z \cap G') \to \operatorname{Hom}(H_2(G/Z;\mathbb{Z}),Z \cap G') \to 0$$

Because we know the right map is surjective, we can find an element of $$H^2(G/Z;Z \cap G')$$ that maps to the element obtained above in $$\operatorname{Hom}(H_2(G/Z;\mathbb{Z}),Z \cap G')$$. Denote the corresponding extension group by $$H$$. We claim that:


 * $$H$$ is isoclinic to $$G$$: In fact, the central subgroup for the extension used to construct $$H$$ is precisely the center of $$H$$.
 * $$Z(H) \le H'$$: We know that the image of the mapping in $$\operatorname{Hom}(H_2(G/Z;\mathbb{Z}),Z \cap G')$$ is surjective to the central subgroup, and also that the image is contained in the derived subgroup, so $$Z(H) \le H'$$.

Journal references

 * : The assertion is made without an explicit proof, but the reasoning for the proof is given in preceding paragraphs, with the backdrop assumption of finiteness.