Why aren't $C_2 \times C_3$ and $D_6$ isomorphic? I am new to group theory so please forgive me for my stupid question. Why are the groups $C_2\times C_3$ not the same as $D_6$? Aren't they both generated by 2 elements one of order 2 the other 3? Is it because for the former the elements are ordered pairs but the latter isn't? And $(a,b)$ in the former means there isn't a $(b,a)$ whereas for $D_6$ we can have $ab$ and $ba$?
 A: $C_2\times C_3$ has an element of order 6; $D_6$ doesn't. 
$C_2\times C_3$ is abelian; $D_6$ isn't. 
A: $C_2 \times C_3$ is an abelian group, whereas $D_6$ is not abelian. in $C_2 \times C_3$, we have $(a,e)(e,b) = (a,b) = (e,b)(a,e)$, but in $D_6$, $ab = b^2a \neq ba$
to go into a little more depth, the generators $(a,e),(e,b)$ in $C_2 \times C_3$ interact differently with each other than the generators $a,b$ in $D_6$. one can think of $C_2 \times C_3$ as "all rotation (in the same direction)" In fact it is isomorphic to the cyclic group of order 6, whereas $D_6$ is "rotations and flips (reflections)". the reflections effectively reverse the direction of the rotations.
there is a notion of such a "twisted direct product" (the semi-direct product), but i am afraid it would just scare you off, if group theory is new to you. the multiplication in it is "complicated".
it's not just enough to have two generators of the same order. the generators have to obey the same relations, too.
A: One of your groups is abelian, while the other is not.
If you take $(a_1, b_1), (a_2, b_2)\in C_2\times C_2$ then you will see that $(a_1, b_1)+(a_2, b_2)=(a_2, b_2)+(a_1, b_1)$. However, the same doesn't hold for $D_6$. If you flip a triangle then rotate it clockwise, this is not the same as rotating it clockwise and then flipping it.
Interestingly, you can think of $D_6$ as ordered pairs. If $a$ is your element of order $3$, $b$ your element of order $2$, then all your elements are of the form $$(a^i, b^j)$$ and they multiply like, $$(a^i, b^j)\cdot(a, b^s)=(a^{i+1}, b^{-j+s})$$ and $$(a^i, b^j)\cdot (1, b^s)=(a^i, b^{j+s}).$$ Notice in the first line the $j$ becomes a negative $j$ but in the second it doesn't. That is, there is some interaction between the first part of the pair and the second.
