# Can invariants distinguish these two 2-groups?

• $G = \{ x,y \mid x^{2^{n+1}} = 1, y^4 = 1, xy = yx^{−1} \}$
• $H = \{ x,y \mid x^{2^{n+1}} = 1, y^4 = 1, xy = yx^{2^n−1}\}$

Can these groups of order $2^{n+3}$ and nilpotency class $n+1$ be distinguished by any reasonable (computable) set of invariants?

While for any set of reasonable invariants, there is surely a pair of groups that share the invariants, it is also surely true that for any pair of groups there is some invariant that distinguishes them.

I personally have not found any significant differences in the conjugacy classes, centralizers, proper subgroups, normalizers, normal subgroups, etc., but perhaps I overlooked something. I've been looking at simple groups lately, and have forgotten how muddy $p$-groups can be.

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I have played around with these two groups for $n=3$ and $n=4$, and for those, one distinction is that the first group has no 2-dimensional irreducible representations with non-real characters (i.e. the Frobenius-Schur indicators of all the 2-dim irreducibles are $\pm1$), while the second has plenty 2-dim irreducible non-real characters. That should be easily provable for all $n$, although I haven't tried yet.

I am not sure whether this can be turned into a purely group theoretic criterion. I briefly tried something in the spirit of this and the answer to this, but to no avail: both groups have the same number of involutions and for both of them, the square roots counting function assumes its maximum at a non-trivial central element.

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Thanks! This is very different from what I was looking at, and is a clear difference. – Jack Schmidt Dec 19 '10 at 4:19

A completely different method of recognition is just a random constructive recognition:

1. Choose a random element $z$
2. If $z$ has order $|G|/4$, call it $x$ [ probability 25% ]
3. If $z$ has order $4$ and if it has more than $2$ conjugates, call it $y$ [ probability 50% ]
4. Otherwise, go to $1$ [ probability 25% ]
5. If we are missing $x$ or $y$, go to $1$.
6. If $x^y = x^{-1}$, return "$G$", else $x^y = x^{(|G|/8)-1}$ and so return "$H$"

This is easy to do since the automorphism groups of $G$ and $H$ act quite transitively on the possible generators. This works well even for large $n$.

This is using the normally "cheating" purely-group-theoretic invariant of "how many isomorphisms to $G$?" In this case, there are just so many, it is easy to find one by random chance.

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