Is this a valid group? Suppose I take $\alpha = e^{\frac{2 \pi i}{3}}$ (the third root of unity) and $\beta = \sqrt{2}$, and the generated set
$$\langle \alpha, \beta | \alpha^3 = e, \beta^2 =e \rangle$$
with the operation being the usual product in the complex numbers. Is this a valid group? I mean, it is closed, associative, has an identity element ($e = 1$) and each element has an inverse (if the element is $\alpha^k\beta^j$, then its inverse is $\alpha^{3-k}\beta^{2-j}$), so it is a group, right? Or did I get one of those wrong?
And if it is a group, would it be correct to say that the only automorphism is the identity mapping? My reasoning being that we must have $\phi(e) = e$, and if we were to have $\phi(\beta) = \beta^k\alpha^j, k=0,1, j =1,2$, then $\phi(\beta^2) = \phi(e) = e \neq \phi(\beta)\phi(\beta) = \beta^{0}\alpha^m, m=1,2$, a contradiction.
I appreciate any corrections! Thanks.
Edits: If the first few comments don't seem to make sense, it's because I had a few things wrong in the original question.
 A: Let us consider the group generated by $e^{2\pi i/3} =: \zeta_3$ and $\sqrt{2}$ with group operation given by multiplication of complex numbers. One might write this group as $G = \langle \zeta_3,\sqrt{2}\rangle\subseteq\Bbb C^\times$. Because $\Bbb C^\times$ (and hence our group $G$) already has a group operation defined on it, we cannot demand that $\sqrt{2}^2 = 1$ without leaving the group we live in, because $\sqrt{2}^2 = 2\neq 1$ inside of $\Bbb C^\times$.
However, group theory gives us a way to start with a group $G$ and then create a new group $G'$ from $G$ such that the desired relationship holds: this is the quotient group construction. In this case, you want the relationship $\sqrt{2}^2 = 1$ to hold (which, as we've noted, does not hold in $\Bbb C$, and hence does not hold in $G$). So what we do is form the quotient of $G$ by the subgroup generated by $2$ inside of $G$. This gives us a new group
$$
G' := G/\langle 2\rangle = \{\zeta_3^a\sqrt{2}^b\langle 2\rangle\mid a,b\in\Bbb Z\},
$$
whose elements are cosets $g\langle 2\rangle = \{g x\mid x\in\langle 2\rangle\} = \{g 2^n\mid n\in\Bbb Z\}$, where $g\in G$. In $G'$, multiplication is performed in the following way: $g\langle 2\rangle\cdot h\langle 2\rangle := gh\langle 2\rangle$ (one must check that this is well-defined, or independent of the choice of representatives $g$ and $h$ used to represent the cosets), and $g\langle 2\rangle$ is the identity of $G'$ if and only if $g\in \langle 2\rangle$.
The quotient group $G'$ admits a natural map from $G$ given by
\begin{align*}
G&\to G'\\
g&\mapsto g\langle 2\rangle
\end{align*}
which is surjective with kernel precisely $\langle 2\rangle$. Note that however, $G'$ is not naturally a subgroup of $G$ or even $\Bbb C$: subgroups map into the given group $G$, but this has a map from our given group $G$ to it.
In general, one can create a group generated by any set $S = \{s_\alpha\mid\alpha\in A\}$ (called generators) subject to the restriction that certain identities $R$ hold within the group. To do this, we begin with the free group $F_S$ on our set of generators $S$, which is the group whose elements are formal strings
$$
s_{\alpha_1}^{n_{\alpha_1}}s_{\alpha_2}^{n_{\alpha_2}}\cdots s_{\alpha_r}^{n_{\alpha_r}},
$$
where each $\alpha_i\in A$, and each $n_{\alpha_i}\in\Bbb Z$, and whose group operation concatenation of strings (just smash them together). Within this group, you may only make simplifications of the form $s_x^{n}\cdot s_x^{-n} = e$. The group has no relationships between the various $s_\alpha$'s. But we may now demand that relationships hold between the $s_\alpha$'s similar to the way we did with $G$: we write our relations in the form $s_{\alpha_1}^{n_{\alpha_1}}s_{\alpha_2}^{n_{\alpha_2}}\cdots s_{\alpha_r}^{n_{\alpha_r}} = e$ (above, you had $\sqrt{2}^2 = 1$), and then we form the quotient group $F/R$ of $F$ by $R$, where $R$ is the normal subgroup of $F$ generated by the relations: i.e., the normal subgroup generated by all the $s_{\alpha_1}^{n_{\alpha_1}}s_{\alpha_2}^{n_{\alpha_2}}\cdots s_{\alpha_r}^{n_{\alpha_r}}$'s that we wanted to be equal to the identity. This group will be the group with the least amount of relationships possible between the elements of $s$, subject to the condition that all of your demanded relationships hold.
We can construct your group $G'$ completely abstractly in this way: we want a group generated by 2 elements $\alpha$ and $\beta$, subject to the conditions that


*

*multiplication is commutative,

*$\alpha^3 = e$, and

*$\beta^2 = e$.


Condition 1 could be expressed as a relation (how?), but we may also simply use it to simplify our problem: we do not need to take the usual free group on two generators $F_2$, but instead, we can start with the free abelian group on two generators $\Bbb Z\alpha\oplus\Bbb Z\beta$, which is much simpler (in particular, every subgroup of an abelian group is normal, so we don't need to worry about that particular point). Now, we want to take the quotient of $\Bbb Z\alpha\oplus\Bbb Z\beta$ by the subgroup generated by $(3\alpha,0)$ and $(0,2\beta)$ (I'm now writing this group additively). This subgroup is simply $3\Bbb Z\alpha\oplus2\Bbb Z\beta$, and when we form the quotient group, we get
$$
(\Bbb Z\alpha\oplus\Bbb Z\beta)/(3\Bbb Z\alpha\oplus2\Bbb Z\beta)\cong (\Bbb Z/3\Bbb Z)\oplus(\Bbb Z/2\Bbb Z)\cong\Bbb Z/6\Bbb Z,
$$
which you can check is isomorphic to the group $G'$.
