# If $\operatorname{class}(G) = 2$ and $\exp(G) = 4$ then $\exp(G') = 2$?

Let $$G$$ be a finite $$p$$-group. I'd like to prove (or disprove) that if the nilpotency class of $$G$$ equals two (i.e., $$1 \neq G' \le Z$$, where $$Z$$ is the center of $$G$$) and the exponent of $$G$$ equals four (i.e., $$g^4 = 1$$ for all $$g \in G$$), then the exponent of $$G'$$ equals two.

Note: By a formula $$[gh, k] = [g, k]^h[h, k]$$, we have $$[gh, k] = [g, k][h, k]$$ because $$G' \le Z$$. So $$[g^2, h] = [g, h]^2$$ in particular and our goal is equivalent to $$G^2 \le Z$$, where $$G^{p^i} := \langle\, g^{p^i} \mid g \in G \,\rangle$$.

I tried to prove it by showing $$(G')^2 \le G^4$$ but I couldn't prove it so far. As $$G' \le G^2$$ is true by a trick $$[g, h] = (g^{-1})^2(gh^{-1})^2h^2$$, I've been expecting something similar does the job. I also checked that the statement is true for $$p$$-groups of order dividing $$2^6$$ by a computer.

• What code & software did you use to check the statement? – Shaun Jan 23 at 12:07
• @Shaun I used GAP. A code is a fairly simple for-loop test using AllSmallGroups. – Orat Jan 23 at 13:24

We have $$[g,h]^2 = (g^{-1}h^{-1}gh)^2 = (g^{-2}gh^{-2}hgh)^2.$$ Now collecting the terms $$g^{-2}$$ and $$h^{-2}$$ and the resulting commutators (which are central) to the left gives $$g^{-2}h^{-2}g^{-2}h^{-2}[g,h^{-2}][h^2,g^{-2}][g^3,h^{-2}](gh)^4 = g^{-2}h^{-2}g^{-2}h^{-2}[g,h^{-2}][h^2,g^{-2}][g^3,h^{-2}].$$ But commutators are bilinear in groups of class $$2$$, and $$[g^2,h^2]=[g,h]^4=1$$, so this simplifies to $$g^{-4}h^{-4}[g^4,h^{-2}][h^2,g^{-2}]=1$$.
• Thank you for your answer. But I'm afraid I don't understand how $(g^{-2}gh^{-2}hgh)^2$ leads to $g^{-2}h^{-2}g^{-2}h^{-2}[g,h^{-2}][h^2,g^{-2}][g^3,h^{-2}](gh)^4$. Could you elaborate it a little more, please? – Orat Jan 23 at 9:39
• This is a standard process known as collection to the left. It's just repeated use of the identity $ba=ab[b,a]$, so for example, the collection process starts on the left of the word with $g^{-2}gh^{-2}h\cdots = g^{-2}h^{-2}g[g,h^{-2}]h\cdots = g^{-2}h^{-2}[g,h^{-2}]gh\cdots$. – Derek Holt Jan 23 at 9:55