How many groups of order at most $25$ are "pleasant" (abelian, with every non-identity element having prime order)?

Notice that by the fundamental theorem of finite abelian groups, we have that if $\Gamma$ is abelian and finite that $$\Gamma \cong \bigoplus_{i=1}^n \mathbb{Z}_{p_i^{e_i}}^{k_i} $$ where each of the $p_i$ are primes. Now, if any of the $e_i$ were bigger than one, so $\mathbb{Z}_{p^\alpha}$ is a subgroup of $\Gamma$ with $\alpha$ greater than 1, it follows that the generator of the cyclic group would have order $p^\alpha$ and therefore not have prime order. It follows that for any pleasant group $\Gamma$ we have $$\Gamma \cong \bigoplus_{i=1}^n \mathbb{Z}_{p_i}^{k_i}$$ Now, suppose that this decomposition is composed of at least two different primes $p_1$ and $p_2$, not equal to each other. Then, by properties of the direct sum of groups $\mathbb{Z}_{p_1} \oplus \mathbb{Z}_{p_2}$ is a subgroup of $\Gamma$, and the element that is the product of the generators of these two groups $\mathbb{Z}_{p_1}$ and $\mathbb{Z}_{p_2}$ would have order $p_1p_2$, and therefore cannot be prime. Thus we have for a pleasant group $\Gamma$ $$\Gamma \cong \mathbb{Z}_{p}^k $$ for $p$ prime and $k$ an integer. Furthermore, it can easily be seen each such group is abelian and that the order of each non-identity element in such a group is $p$. Thus pleasant groups are in 1-1 correspondence with prime powers. Therefore, the number of pleasant groups of order at most 25 is simply equal to the number of prime powers less than 25. There are exactly 15 such numbers $$ 1, 2, 3, 4, 5, 7, 8, 9, 11, 13, 16, 17, 19, 23, 25$$ which correspond to precisely the groups you have found. And thus it seems that there are 15 pleasant groups of order at most 25. I could be mistaken somewhere in this reasoning but it seems to me that the answers given do not contain the correct choice of 15.