Order-cum-power statistics-equivalent not implies 1-isomorphic

History

This question was answered on Math Overflow here.

Statement

It is possible to have two groups $G$ and $H$ that are order-cum-power statistics-equivalent (i.e., they have the same order-cum-power statistics) but are not 1-isomorphic.

Proof

Further information: direct product of Q8 and Z4, SmallGroup(32,35)

We construct groups $G$ and $H$ that are both of order $32=2^{5}$:

1. $G$ is defined as the direct product of Q8 and Z4, i.e., it is the external direct product of the quaternion group and cyclic group of order 4.
2. $H$ is defined as SmallGroup(32,35). One way of describing this group is as a maximal subgroup of direct product of Q8 and Q8 that is not isomorphic to direct product of Q8 and Z4.

Both $G$ and $H$ are groups of order $32$. They both have 1 identity element, 3 non-identity elements of order 2 that are squares, and 28 elements of order 4 that are not proper powers. Thus, they have the same order-cum-power statistics.

However, $G$ is not 1-isomorphic to $H$. To see this, note that the three non-identity elements of $G$ have $12$, $12$, and $4$ square roots. On the other hand, the three non-identity elements of $H$ have $20$, $4$ and $4$ square roots. Since the number of square roots of an element must be preserved under a 1-isomorphism, $G$ and $H$ are 1-isomorphic.

References

• Math Overflow question whose answer contained this example