I believe I've found a third and fourth group of degree four. Tell me how wrong I am. According to every source I could find, the Cyclic group Z4 and the Klein four-group are the only two groups of degree 4.  As I'm new to group theory, I decided to figure out the groups by myself by creating multiplication tables through simple logical steps.  It was a trivial process for degrees 1,2,and 3, but when I got to degree 4 I found "new groups".  I've been trying to figure out where I went wrong, so I'm going to give you the multiplication tables of my "new groups" and hope that someone will tell me why they're not actually groups.  (I also found a couple groups of degree 5, and from what I know, there is a rule about groups of prime degree that say that I should be wrong, but I don't know why the extra ones that I found aren't groups either.  I'm hoping that these responses will help clear that up for me.)

$$\begin{matrix}1&2&3&4\\
2&1&4&3\\
3&4&2&1\\
4&3&1&2\end{matrix}$$
and:
$$\begin{matrix}1&2&3&4\\
2&4&1&3\\
3&1&4&2\\
4&3&2&1\end{matrix}$$
Thank you for any help.
 A: Let's call your first group $G = \{e,a,b,c\}$, and your second group $H = \{1,x,y,z\}$.
Note that $b^2 = a$, so we may re-write $G = \{e,b^2,b,c\}$. Next, note $b^3 = b^2b = ab = c$, so we have $G = \{e,b^2,b,b^3\} = \{e,b,b^2,b^3\}$ making it clear that $b$ generates $G$.
This suggests the map $\Bbb Z_4 \to G$:
$0 \mapsto e\\1 \mapsto b\\2 \mapsto b^2 = a\\3 \mapsto b^3 = c$
and to verify this is an isomorphism, you need only verify:
$k + m(\text{ mod }4) \mapsto b^{k+m(\text{ mod }4)} = b^kb^m$ for $k, m \in \{0,1,2,3\}$
A similar analysis holds with $H$ using $x$ as a generator.
You're correct in some sense in that your groups are indeed distinct products on a set of four elements, but "re-naming" the elements leads to the same kind of group structure (cyclic) which is what the notion of isomorphism is meant to capture (the same in all important algebraic ways, but not necessarily identical sets).
A: They are both isomorphic to $\Bbb Z_4$.  The key is to find the element of order $2$, so the correspondence is $$\begin {array}{c|c|c} \Bbb Z_4 & 1 & 2 \\ \hline 0 &1&1\\1&3&3\\2&2&4\\3&4&2\end {array}$$  As $\Bbb Z_4$ is isomorphic under the interchange $1 \leftrightarrow 3$ you can make that swap here if you want.
