# 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.

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What have you tried? –  Tobias Kildetoft Jan 16 '13 at 20:49
oeis.org/A006116 –  Hagen von Eitzen Jan 16 '13 at 20:56
I tried to do $P\times P$ first and the elements are $(e,e),(e,a),(a,e),(a,a)$, right? –  bbr4in Jan 16 '13 at 21:02
That's a good start. Your list of the elements is correct, but what are the subgroups of $P \times P$? –  Hew Wolff Jan 16 '13 at 21:30
It's good to keep in mind that there is only one group of order two, so you may always think of that one! –  andybenji Jan 16 '13 at 21:35

$P = \{e, a\} \cong \{0, 1\} = \mathbb{Z}_2$. 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 $4$.

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Start first by computing the subgroups of $P\times P$. One, of course, is $\{(e, e)\}$. By the definition of the direct product, if the operation on $P$ is $*$, then compute $(e, a)(e, a) = (e*e, a*a) = (e, e)$. (Recall, a is of order 2, so $a*a = e$). So what can you say about $\{(e, e), (e, a)\}$?...etc. –  amWhy Jan 16 '13 at 22:08
There are certainly subgroups of order 4. –  Gerry Myerson Jan 16 '13 at 23:13
I count 16 subgroups. –  Gerry Myerson Jan 16 '13 at 23:31
@user52187: feel free to follow up with any questions you still have... –  amWhy Jan 17 '13 at 22:00