I didn't understand how to write down the alternating group A3. Is this the group consisting of only the even permutations? Also, what familiar group is this isomorphic to?
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$\begingroup$Yes, $A_3$ is the set of all even permutations in $S_3 = \{id, (12), (13), (23), (123), (132)\}$.
Remember that an even permutation can be written as the product of an even number of transpositions. The identity of any symmetric group is even, because id can be written as the product of two transposition. In this case, e.g. $id = (1,2)(2, 1)$.
Note also that $(123) = (12)(23)$, and $(132) = (13)(32).$ So, $$A_3 = \{id, (123), (132)\}.$$
Since $|A_3| = 3$, and the fact that there is only one group, up to isomorphism, of order $3$, $$A_3 \cong \mathbb Z_3,$$ where $\mathbb Z_3$ is the cyclic group under addition modulo $3$.
$\endgroup$ 2 $\begingroup$You can also have an isomorphism with the following matrix group:
$$I=\begin{pmatrix}1&0&0\\0&1&0\\0&0&1\end{pmatrix}, \ \ C=\begin{pmatrix}0&0&1\\1&0&0\\0&1&0\end{pmatrix}, \ \ C^2=\begin{pmatrix}0&1&0\\0&0&1\\1&0&0\end{pmatrix}$$
Explanation: if $e_1,e_2,e_3$ is the canonical basis
$I$ will sent all $e_i$ on themselves,
$C$ will send $e_1$ onto $e_2$, $e_2$ onto $e_3$, $e_3$ onto $e_1$, that's a cycle !
$C^2$ will... be as well $C^{-1}$, because $C^3=I.$
Its elements are the $3$-cycles and the identity
$\begin{cases} f(1)=2, f(2)=3, f(3)=1 \\ g(1)=3, g(2)=1, g(3)=2 \\ Id \end{cases}$.
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