# Intermediately characteristic not implies isomorph-containing in group of prime power order

This article gives the statement and possibly, proof, of a non-implication relation between two subgroup properties, when the big group is a group of prime power order. That is, it states that in a group of prime power order, every subgroup satisfying the first subgroup property (i.e., intermediately characteristic subgroup) need not satisfy the second subgroup property (i.e., isomorph-containing subgroup)
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This article gives the statement and possibly, proof, of a non-implication relation between two subgroup properties. That is, it states that every subgroup satisfying the first subgroup property (i.e., characteristic maximal subgroup of group of prime power order) need not satisfy the second subgroup property (i.e., isomorph-free subgroup of group of prime power order)
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## Statement

For any prime number $p$, there exists a finite $p$-group $G$ and a subgroup $H$ of $G$ such that $H$ is an intermediately characteristic subgroup of $G$ but is not an isomorph-containing subgroup (and hence not an Isomorph-free subgroup (?), or Isomorph-free subgroup of group of prime power order (?)) of $G$.

In fact, we can obtain an example where $H$ is a characteristic maximal subgroup of $G$. Thus, a characteristic maximal subgroup of a group of prime power order need not be an Isomorph-free maximal subgroup of group of prime power order (?).

## Proof

Further information: SmallGroup(16,4), subgroup structure of SmallGroup(16,4)

For any $p$, set $G$ to be SmallGroup(p^4,4) and $H$ to be the unique characteristic subgroup of order $p^3$. $H$ is a direct product of a cyclic group of order $p^2$ and a cyclic group of order $p$. All the other maximal subgroups of $G$ are also isomorphic to $H$.

For instance, for $p = 2$, set $G$ to be SmallGroup(16,4) and $H$ to be the unique characteristic subgroup of order eight. Explicitly: $G := \langle a,b,c \mid a^2 = b^4 = e, b^2 = c^2, ab = ba, ac = ca, cbc^{-1} = ab \rangle, \qquad H = \langle b^2, bc \rangle$.

Both the other subgroups of order eight are isomorphic to $H$, and all are isomorphic to the direct product of Z4 and Z2.