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 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 $W=Z\cap G^{\prime }$. 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.

$0\to operatorname{E}^x((G/Z{)}^{\mathrm{a}\mathrm{b}};Z)\to H^2(G/Z;Z)\to \mathrm{H}\mathrm{o}\mathrm{m}(H_2(G/Z;\mathbb{Z}),Z)\to 0$

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 $\mathrm{H}\mathrm{o}\mathrm{m}(H_2(G/Z;mathbZ),Z)$. It turns out from the constriction (see the formula page) that the image of the mapping must necessarily lie in the subgroup $Z\cap G^{\prime }$. In fact, the image of the homomorphism is precisely $Z\cap G^{\prime }$. Thus, in fact, we get an element of $\mathrm{H}\mathrm{o}\mathrm{m}(H_2(G/Z;\mathbb{Z}),Z\cap G^{\prime })$.

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 $Z\cap G^{\prime }$ used in place of $Z$ as the central subgroup:

$0\to operatorname{E}^x((G/Z{)}^{\mathrm{a}\mathrm{b}};Z\cap G^{\prime })\to H^2(G/Z;Z\cap G^{\prime })\to \mathrm{H}\mathrm{o}\mathrm{m}(H_2(G/Z;\mathbb{Z}),Z\cap G^{\prime })\to 0$

Because we know the right map is surjective, we can find an element of $H^2(G/Z;Z\cap G^{\prime })$ that maps to the element obtained above in $\mathrm{H}\mathrm{o}\mathrm{m}(H_2(G/Z;\mathbb{Z}),Z\cap G^{\prime })$. 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 $H$.
• $Z(H)\le H^{\prime }$: We know that the image of the mapping in $\mathrm{H}\mathrm{o}\mathrm{m}(H_2(G/Z;\mathbb{Z}),Z\cap G^{\prime })$ is surjective to the central subgroup, and also that the image is contained in the derived subgroup, so $Z(H)\le H^{\prime }$.

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.