Subgroups of $C_2\times C_6$ WolframMathWorld says there are $10$ subgroups. I can find $8$:
$$
\{1\}\times\{1\}\\
\{1\}\times C_2\\
C_2\times\{1\}\\
\{1\}\times C_3\\
C_2\times C_2\\
\{1\}\times C_6\\
C_2\times C_3\\
C_2\times C_6
$$
And this answer says that the subgroups of a direct product of cyclic groups are precisely the direct product of the individual subgroups. Why does Wolfram say there are $10$? What am I missing?
 A: Yes, it has more groups than those. If $C_2=\{e,g\}$ and $C_6=\{e,h,h^2,h^3,h^4,h^5\}$, then, for instance, $\{(e,e),(g,h^3)\}$ is another subgroup of $C_2\times C_6$ (note that $g$ and $h^3$ are the elements of order $2$ of $C_2$ and of $C_6$ respectively).
A: The answer you linked is answering a slightly different question.
Just look to the Klein four group for an example, the diagonal subgroup, of a subgroup that is not a direct product of subgroups.  That's the subgroup $\langle(1,1)\rangle$.
You missed a spot.
A: It is not true that a subgroup of $A \times B$ has the form $C \times D$ where $C \leq A$ and $D \leq B$. Not in general and not even if $A$ and $B$ are cyclic. 
As an example where this fails you can take $G = C_2 \times C_2$. Apart from the whole group and the trivial subgroup, $G$ has $3$ subgroups of order $2$. One is the first copy of $C_2$, the other is the second copy, and the third is the diagonal subgroup (and it is precisely that subgroup that does not admit the direct decomposition referred to above).
But it is true that if $\gcd(|A|,|B|)=1$ then all subgroups of $A \times B$ are subproducts, i.e. of type $C \times D$ where $C \leq A$ and $D \leq B$. Now that you know the five subgroups of $C_2 \times C_2$ and you of course know the two subgroups of $C_3$, just observe that your group $$C_2 \times C_6 \cong (C_2 \times C_2) \times C_3$$
and use the coprimality theorem I mentioned.
