Is any finite indecomposable group of exponent $4$ isomorphic either to $C_4$ or to $Q_8$?

Question

Suppose $G$ is a finite group of exact exponent $4$, which is not decomposable into semidirect product. Is it true, that $G$ is isomorphic either to $C_4$ or to $Q_8$?

It is true for groups of order $4$ as $C_4$ is the only such group.

It is true for groups of order $8$ as only $C_4 \times C_2$, $C_4 \rtimes C_2$ and $Q_8$ have exact exponent $4$.

It is true for groups of order $16$ as only $C_4 \times C_2 \times C_2$, $C_2 \times (C_4 \rtimes C_2)$, $C_2 \times Q_8$, $C_4 \times C_4$, $C_4 \rtimes C_4$, $(C_2 \times C_2) \rtimes C_4$ and $(C_4 \times C_2) \rtimes C_2$ have exact exponent $4$.

Do not know, however, whether it is true in general or not...

Answer 1

There are further examples. There is one of order 32, $\mathtt{SmallGroup}(32,32)$, which has the presentation:

 G.1^2 = G.4, 
 G.2^2 = G.5, 
 G.3^2 = G.4, 
 G.2^G.1 = G.2 * G.4, 
 G.3^G.1 = G.3 * G.5

with the convention that generators have order 2, and two generators commute unless otherwise specified.

There are four examples of order 64, namely the groups with small group numbers 79, 81, 82, 245, and lots more of order 128, so it seems unlikely that a classification is feasible.