Elementary abelian-to-normal replacement fails for Klein four-group

This article discusses a failure of replacement, i.e., a situation where the analogue of a valid replacement theorem fails to hold under slightly modified conditions.
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Statement

Hands-on statement

We can find a finite group whose order is a power of $2$, that contains a Klein four-group (?) as a subgroup(i.e., it contains an elementary abelian subgroup of order four) but does not contain any normal subgroup that is a Klein four-group.

In fact, the following stronger statement is true: for any $k$, we can find a finite group whose order is a power of $2$, that contains a Klein four-subgroup but does not contain a $k$-subnormal Klein four-subgroup.

Statement in terms of the weak normal replacement condition

The single-element collection $\mathcal{S}$ comprising the Klein four-group is not a Collection of groups satisfying a weak normal replacement condition (?) for the prime $2$.

Related facts

Similar facts

• Collection of groups of prime-to-the-prime order and prime exponent does not satisfy weak normal replacement condition

Opposite facts

• Elementary abelian-to-normal replacement theorem for prime-square order: The statement that fails for $2$ is true for odd primes.
• Abelian-to-normal replacement theorem for prime-cube order
• Jonah-Konvisser elementary abelian-to-normal replacement theorem
• Jonah-Konvisser abelian-to-normal replacement theorem

Proof

Example of the dihedral group (for the weaker version)

Further information: dihedral group:D16, Subgroup structure of dihedral group:D16

Consider the dihedral group of order sixteen:

$G := \langle a,x \mid a^8 = x^2 = 1, xax = a^{-1} \rangle$.

Then:

• The subgroup $\langle a^4, x \rangle$ is a Klein four-group.
• There is no normal Klein four-subgroup: Any subgroup contained in $\langle a \rangle$ is cyclic, so it cannot be a Klein four-group. Thus, a normal Klein four-subgroup, if it exists, must contain some element outside $\langle a \rangle$. However, the sizes of the conjugacy classes of elements outside $\langle a \rangle$ is $4$ each, so any normal subgroup containing an element outside $\langle a \rangle$ must have order strictly bigger than four.

Example of the dihedral group (for the stronger version)

Further information: Dihedral group, Subgroup structure of dihedral groups

We need to take a dihedral group of order $2^{k+3}$, as follows:

$G := \langle a,x \mid a^{2^{k+2}} = x^2 = 1, xax = a^{-1} \rangle$.

Then:

• The subgroup $\langle a^{2^{k+1}},x \rangle$ is a Klein four-group.
• There is no $k$-subnormal Klein four-subgroup: Any subgroup contained in $\langle a \rangle$ is cyclic, so it cannot be a Klein four-group. Thus, a $k$-subnormal Klein four-subgroup, if it exists, must contain some element outside $\langle a \rangle$. Since all elements outside $\langle a \rangle$ are in the same automorphism class, we can assume that it contains $x$. However, the smallest $k$-subnormal subgroup containing $x$, computed by taking the normal closure $k$ times, is a dihedral group of order eight given by $\langle a^{2^k},x \rangle$.