Does $D_4$ have a verbal subgroup of order 4?

How did this question arise:

In the comments $Q_8$ ad $D_4$ were pointed to be a possible counterexample to this question: Is it true, that for any two non-isomorphic finite groups $G$ and $H$ there exists such a group word $w$, that $|V_w(G)| \neq |V_w(H)|$?

However, $Q_8$ does not have a characteristic subgroup of order $4$, whereas $D_4$ has $3$ normal subgroups of order $4$. So, if one of those subgroups happens to be verbal for some group word $w$, then $Q_8$ ad $D_4$ are definitely not a counterexample.

If such $w$ exists, then it is clearly not an identity in $D_4$. Also, $w \neq [x, y]$, as $D_4’ \cong C_2$, $w \neq x$, as $V_{x^3}(D_4) \cong D_4$, $w \neq x^2$ as $V_{x^2}(D_4) \cong C_2$ and $w \neq x^3$, as $V_{x^3}(D_4) \cong D_4$.

However, I do not know, how to proceed further.

  • $\begingroup$ Is $Q_8$ the quaternion group of order $8$? That has a few normal subgroups of order $4$. $\endgroup$ – Matt Samuel Feb 16 at 19:42
  • $\begingroup$ @MattSamuel, that was a typo. I meant "characteristic subgroups", when I talked about $Q_8$. $\endgroup$ – Yanior Weg Feb 16 at 20:27

The answer is apparently no: none of the three subgroups of $D_4$ of order $4$ are verbal.

Incidentally, any subgroup of order four in $Q_8$ is normal (since of index $2$).


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