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

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.

• Is $Q_8$ the quaternion group of order $8$? That has a few normal subgroups of order $4$. – Matt Samuel Feb 16 at 19:42
• @MattSamuel, that was a typo. I meant "characteristic subgroups", when I talked about $Q_8$. – 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 $$2$$).