Direct product of group order 2 If $P$ is group of order 2, how many subgroups (trivial and proper) has the group $P \times P \times P$? Labelling the elements of $P$ to be $e$ and $a$, list the proper subgroups. 
 A: $P = \{e, a\} \cong \{0, 1\} = \mathbb{Z}_2$. There is one and only one group of order $2$ and it is cyclic. So $e$ is the identity, $a^2 = aa = e$.
Yes, your computation of $P\times P = \{(e,e),(e,a),(a,e),(a,a)\}$ is correct. $P\times P$ has 3 subgroups of order 2, the trivial subgroup $\{e, e\},$ and $P\times P$ itself: $5$ in all, 4 not including the improper subset $P\times P$.
Can you try and compute $(P\times P) \times P = P\times P \times P$? The order of $P\times P \times P$ is $8$, and it consists of 3-tuples, e.g., $(e, e, e)$ is the identity. Hint: append each of the elements you computed for $P \times P$ first with $e$, then append the same element from $P\times P$ with $a$: e.g. $(e, e) \to (e, e, e), (e, e, a); (e, a) \to (e, a, e), (e, a, a)$...
Your task is to determine the subgroups (proper and trivial) in this group of eight 3-tuples:
Hint: there are 15 subgroups, excluding $P\times P \times P$, (so 16 in all). Can you find the proper subgroups? One of order $1$: the identity, the rest are of order $2$ or $4$: use the definition of direct product to compute which elements form subgroups of order 2, and which comprise subgroups of order 4, and confirm that there are no other subgroups than those of order $1$ and $2$ and $4$.
