Why is $\langle \mathbb{Z}_4, + \rangle$ not isomorphic to $\langle \mathbb{Z}_2 \times \mathbb{Z}_2, + \rangle$? I'm having some trouble here, specifically with the idea of $\langle \mathbb{Z}_2 \times \mathbb{Z}_2, + \rangle$ as a group. Can anyone help me out with some explanations? 
Moreover, I generally haven't wrapped my head around groups like that. Can anyone shed some light on exactly how to think about, for example, $\langle \mathbb{Z}_2 \times \mathbb{Z}_3, + \rangle$?
 A: Think of an element of $\Bbb Z_n \times \Bbb Z_m$ as having two parts, written $\langle p,q\rangle$. The left part $p$ is an element of $\Bbb Z_n$ and the right part  $q$ is an element of $\Bbb Z_m$.  The two parts never interact.  To add $\langle a, b\rangle$ to 
$\langle c, d\rangle$, you add the two parts separately, and get
$\langle a+c, b+d\rangle$.  The left-side addition is done $\Bbb Z_n$-style, because $a$ and $c$ are elements of $\Bbb Z_n$, and the right-side addition is done $\Bbb Z_m$-style.
$\def\V{\Bbb Z_2\times \Bbb Z_2}\V$ has four elements, which are $\langle 0,0\rangle,
\langle 0,1\rangle,
\langle 1,0\rangle,$ and 
$\langle 1,1\rangle$. Its identity element is $\langle 0,0\rangle$.  But $\V$ is not $\Bbb Z_4$, and the easiest way to see this is 
that every element $x$ of $\V$ has the property  that $x+x$ is the identity element $\langle 0,0\rangle$:
$$\begin{align} 
\langle 0,0\rangle + \langle 0,0\rangle  & = \langle 0,0\rangle\\
\langle 0,1\rangle + \langle 0,1\rangle  & = \langle 0,0\rangle\\
\langle 1,0\rangle + \langle 1,0\rangle  & = \langle 0,0\rangle\\
\langle 1,1\rangle + \langle 1,1\rangle  & = \langle 0,0\rangle
\end{align}
$$
all the additions being done $\Bbb Z_2$-style.
But $\Bbb Z_4$ is different; it has two elements, 1 and 3, that do not have the property that $x+x=0$. They have $1+1 = 3+3 = 2$ instead, and in $\Bbb Z_4, 2\ne 0$.  $\V$ has nothing like this.
Or looked at in the opposite direction, $\Bbb Z_4$ has an operation, namely the operation of adding 1, which you must do four times before you get back to where you started; it is analogous to giving an object a quarter turn. After four quarter turns, and no fewer, the object has returned to its original position. $\V$ has nothing like this; every operation in $\V$ gets you back to where you started after at no more than two repetitions.
A: $\mathbb Z_2\times\mathbb Z_2$ is the set of ordered pairs $(a,b)$ with $a,b\in\mathbb Z_2$, i.e. $\{(0,0),(1,0),(0,1),(1,1)\}$. The group operation is coordinate-wise $\mathbb Z_2$-addition. 
It cannot be isomorphic to $\mathbb Z_4$ because all nonzero elements in $\mathbb Z_2\times\mathbb Z_2$ have order 2, while in $\mathbb Z_4$ there are elements of order 4. In particular, $\mathbb Z_2\times\mathbb Z_2$ is not cyclic. 
A: These finite groups are great because you can list all of their elements.
For instance, here are the elements of $\mathbb{Z}_4$: $\{0,1,2,3\}$.
And here those of $\mathbb{Z}_2\times \mathbb{Z}_2$: $\{ (0,0), (0,1), (1,0), (1,1)\}$.
(I'm omitting in both cases to put some $\overline{2}$ all above those numbers in order to point out that they are not actual natural numbers, but classes. But this way I can write faster.)
So, this group $\mathbb{Z}_4$ has a particular guy, $1$, inside with this property:
$$
1 + 1 = 2, \ 1 + 1 + 1 = 3, \ 1 + 1 + 1 + 1 = 4 = 0 \ .
$$
Can you find anyone inside $\mathbb{Z}_2\times \mathbb{Z}_2$ with a similar behaviour? Namely, that four times itself produces the neutral element? [EDIT: Thanks to MJD's remark. And not just two times?] Nope? You can't? -Then, they cannot be isomorphic.
A: Recall that $$\mathbb Z_{mn} \cong \mathbb Z_m \times \mathbb Z_n\;\;\text{ if and only if }\;\;\gcd(m, n) = 1$$
In your first case, take $m = n = 2.$
In the case of $\mathbb{Z}_2 \times \mathbb{Z}_3,\;$ we DO have that $\mathbb Z_6 = \mathbb Z_{2 \times 3} \cong \mathbb{Z}_2 \times \mathbb{Z}_3,\;$ since $\;\gcd(2, 3) = 1$.
