Variety-containing implies omega subgroup in group of prime power order

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Suppose P is a group of prime power order, i.e., a finite p-group for some prime number p. Suppose H is a Variety-containing subgroup (?) of P: a subgroup of P such that any subgroup of P isomorphic to a subgroup of H is itself contained in H. Then, H is one of the omega subgroups of P. More specifically, if the exponent of H is p^k, then:

H = \Omega_k(P) := \langle x \mid x^{p^k} = e \rangle.

Note that by the equivalence of definitions of variety-containing subgroup of finite group, assuming that H is a variety-containing subgroup of P is equivalent to assuming that it is a subhomomorph-containing subgroup or that it is a subisomorph-containing subgroup.


Given: A finite p-group P, a variety-containing subgroup H of P.

To prove: H = \Omega_k(P) for some natural number k.

Proof: Let p^k be the exponent of H. Then, we clearly have:

H \le \Omega_k(P) = \langle x \mid x^{p^k} = e \rangle.

Next, we show that H = \Omega_k(P). Suppose x \in P is such that x^{p^k} = e. Then, since p^k is the exponent of H, there exists y \in H such that the order of y is p^k. Suppose the order of x is p^l, l \le k. Then, \langle x \rangle \cong \langle y^{p^{k-l}} \rangle. Thus, \langle x \rangle is isomorphic to a subgroup of H, so \langle x \rangle \le H. Thus, \Omega_k(P) = H.