Let A be a cyclic group with prime order p. Then the direct product $A\times A$ should have order $p^{2}$ according to wikipedia.
http://en.wikipedia.org/wiki/Direct_product_of_groups#Elementary_properties
However I found this theorem in the book that states for subgroups H,K, of G that
$ o(HK) = o(H)o(K)/((o(H \cap K) ) $
But $A$ is a subgroup of $A$ and $A \cap A = A$ so then $ o(AA) = o(A)o(A)/o(A) = o(A) $
The last result makes sense to me, as $AA = A$.
Maybe I'm confused about something. The external direct product $A\times A$ is isomorphic to the internal direct product $AA$, but I'm getting different orders for them.
What am I doing wrong ?
Best Answer
There are certain conditions that have to be satisfied for a group to be considered an internal direct product of two subgroups.
Let $G$ be a group and let let $H$, $K$ be subgroups. Then $G$ is the internal direct product of $H$ and $K$ if $G=HK$, $H\cap K =1$ and $hk=kh$ for all $k\in K$, $h\in H$. In this case your order formula gives the correct order. In your example $A\cap A$ has nontrivial intersection so it cannot be considered an internal direct product. The idea here is that every element can be expressed uniquely in the form $g=hk$ and $(h_1k_1)(h_2k_2)=(h_1h_2)(k_1k_2)$, which makes the isomorphism to the external direct product much clearer.
In general multiplying two groups may not even be a group, though if one is normal it will be, but it still may not be a direct product. Look up semidirect product