$\mathbb{Z}_6/\mathbb{Z}_2$ isomorphic to $\mathbb{Z}_3$? Recently in class my teacher mentioned that the quotient group $\mathbb{Z}_6/\mathbb{Z}_2$ is isomorphic to $\mathbb{Z}_3$.
May I ask why is this so?
Also, what do elements in $\mathbb{Z}_6/\mathbb{Z}_2$ look like? I thought they were of the form $a+\mathbb{Z}_2$, but then there would be 6 elements ($1\leq a\leq 6$), which is not quite right.
Thanks for any help!
 A: How is $\mathbb Z_2:=\mathbb Z/2\mathbb Z$ a subgroup of $\mathbb Z_6:=\mathbb Z/6 \mathbb Z$ in the first place?
Well, it isn't in a very natural way (at least if $2,6$ were replaced by bigger numbers, we might run into choices). To obtain an injective homomorphism $\mathbb Z_2\to\mathbb Z/6$, we need a homomorphism $\mathbb Z\to\mathbb Z/6\mathbb Z$ that has $2\mathbb Z$ in its kernel. I suggest $x\mapsto 3x+6\mathbb Z$, so that the desired injection $\mathbb Z_2\to\mathbb Z_6$ is $x+2\mathbb Z\mapsto 3x+6\mathbb Z$.
With this homomorhism understood as inclsion you may show that $\mathbb Z_6/\mathbb Z_2\cong \mathbb Z_3$.
A: $|\mathbb{Z}_6/\mathbb{Z}_2| = 6/2 = 3$ and there is only a group of order 3, namely...
A: The subgroup of $\mathbb{Z}_{6}$ that is isomorphic to $\mathbb{Z}_{2}$ contains the elements $0$ and $3$ (since $3+3 = 0$ in $\mathbb{Z}_{6}$).  So consider $\mathbb{Z}_{2} = \{0,3\} \subset \mathbb{Z}_{6}$, then $1+\mathbb{Z}_{2} = \{1,4\} = 4 + \mathbb{Z}_{2}$. This should let you complete the exercise (you are correct about how to represent the elements of the factor group).
A: $\mathbb{Z}_{pq}\cong \mathbb{Z}_{p}\times \mathbb{Z}_{q}$ if and only if $p$ and $q$ are relatively prime.  Let $p=2$ and $q=3$....  Then convince yourself that 
$$\frac{\mathbb{Z}_{p}\times \mathbb{Z}_{q}}{\mathbb{Z}_{p}}\cong \mathbb{Z}_{q}$$
EDIT:  For the first part, go to:http://mathforum.org/library/drmath/view/73205.html
For the centered equation: In general, if you have $G\times H$ then any element from here is of the form $(g,h)$ for $g\in G$ and $h\in H$.  When you "mod-out" by $G$ then you are saying any elements from here are the same.  So for example $(g_1,h)=(g_2,h)$ for any $h\in H$ and any $g_i\in G$.  But then distict elements of this space are distinct if and only if the second coordinate is different (that is, $(a,b)\neq (c,d)$ if and only if $b\neq d$.... but this is precisely $H$).
A: consider the surjective homomorphism from Z6 to Z3 - f(i(mod6))=i(mod3)
(you must show this function is well defined)
and use the fundamental homorphism thm.
