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., $Z(H)\leq [H,H]$.
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 $Z(H)\leq [H,H]$.

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 $H$.
• $Z(H)\leq H'$: 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 $Z(H)\leq H'$.

References

Journal references

• The classification of prime-power groups by Philip Hall, Volume 69, (Year 1937): ^{Official linkMore info}: 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.