Finite abelian groups of order 72 that have elements of order $4$?

Question

We have that $72=2^3\times 3^2$. Since $3=1+1+1 = 2+1$ and $2=1+1$ we have $3\times 2 = 6$ finite abelian groups of order $72.$

I think that $Z/8Z\times Z/9Z$, $Z/8Z\times Z/3Z \times Z/3Z$, $Z/4Z\times Z/2Z\times Z/9Z$ and $Z/4Z\times Z/2Z\times Z/3Z \times Z/3Z$ are the only groups with element of order $4.$

Is this answer correct?

Answer 1

An abelian group of the form $\frac{Z}{m_1Z} \times ... \times \frac Z{m_nZ}$ has an element of order $k$ if and only if $k$ is the lcm of some $d_1,...,d_n$ which are divisors of $m_1,...,m_n$ respectively.

Keeping this in mind, to ensure that $4$ is an LCM of any possible set of divisors, all we need to do is ensure that one of the $m_i$ is a multiple of $4$. From here, it is easy to see that your answer is correct, since we only needed to avoid anything having $\mathbb Z_2 \times \mathbb Z_2 \times \mathbb Z_2$ in its decomposition.

Answer 2

Since $72=2^3 3^2$, there are $p(3)p(2)=3\cdot2=6$ abelian groups of order $72$.

Of the $3$ possibilities for the $2$-Sylow, only $2$ work.

Therefore, there are $2\cdot2=4$ abelian groups of order $72$ having an element of order $4$.