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

Hands-on statement
We can find a finite group whose order is a power of $$2$$, that contains a fact about::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 fact about::collection of groups satisfying a weak normal replacement condition for the prime $$2$$.

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

Example of the dihedral group (for the weaker version)
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)
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 $$\langle a^{2^k},x \rangle$$.