# Isoclinic groups have same nilpotency class

## Statement

Suppose $G_1$ and $G_2$ are Isoclinic groups (?). Then, the following are true:

• $G_1$ is a nilpotent group if and only if $G_2$ is a nilpotent group.
• If the groups are nilpotent and both nontrivial, then the nilpotency class of $G_1$ is the same as the nilpotency class of $G_2$. Note that if one of the groups is trivial, the other may be nontrivial but must still be abelian, giving a situation where one group has class zero and the other has class one.

## Proof

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Given: Isoclinic groups $G_1$ and $G_2$.

To prove: $G_1$ is nilpotent if and only if $G_2$ is, and if so, they have the same nilpotency class if both are nontrivial. If either is trivial, the other may be nontrivial but must be abelian.

Step no. Assertion/construction Facts used Previous steps used
1 $G_1$ is nilpotent if and only if its inner automorphism group is nilpotent, and if so, the nilpotency class of $G_1$ is one more than the nilpotency class of its inner automorphism group (unless $G_1$ is trivial). definition of nilpotent group, via the upper central series. --
2 $G_2$ is nilpotent if and only if its inner automorphism group is nilpotent, and if so, the nilpotency class of $G_2$ is one more than the nilpotency class of its inner automorphism group (unless $G_2$ is trivial). definition of nilpotent group, via the upper central series. --
3 The inner automorphism group of $G_1$ is isomorphic to the inner automorphism group of $G_2$. definition of isoclinism $G_1$ is isoclinic to $G_2$.
4 $G_1$ is nilpotent if and only if $G_2$ is nilpotent, and they have the same nilpotency class unless one of them is trivial. Steps (1)-(3)
5 If either group is trivial, the inner automorphism group of both must be trivial, so both must be abelian. Step (3).

Steps (4) and (5) together complete the proof.