# Presentation of $Gal\left( \mathbb Q(\sqrt{2},i)/\mathbb Q\right)$

Let $E = \mathbb Q(\sqrt{2}, i)$ and $\zeta$ be a primitive root of unity. We have $[E:\mathbb Q] = 16$. The possible elements of $G := Gal\left(E/\mathbb Q\right)$ are:

$$\begin{cases} \sqrt{2} \mapsto \zeta^k \sqrt{2} \\ i \mapsto \pm i \end{cases}: 0\le k\le 7$$

And these are precisely the elements of $G$ because $[E:\mathbb Q]=16$.

Now let:

$$\sigma: \begin{cases} \sqrt{2} \mapsto \zeta \sqrt{2} \\ i \mapsto i \end{cases}, \tau: \begin{cases} \sqrt{2} \mapsto \sqrt{2} \\ i \mapsto -i \end{cases}$$

Then it can be seen that $G = \langle \sigma, \tau \rangle$ by computing $\sigma^k$ and $\tau \sigma^k$, $1\le k \le 7$.

Now $\sigma^8 = \tau^2 = 1$ and $\sigma \tau = \tau \sigma^3$.

My question is: isn't this sufficient to say the following?

$$G = \langle \sigma, \tau \mid \sigma^8 = \tau^2 = 1, \sigma \tau = \tau \sigma^3 \rangle$$

What else needs to be checked?

• Although I didn't check the computations, assuming they are correct, then yes, it is sufficient but to prove that (in fact to prove that any group of order $16$ satisfying the above conditions is a dihedral group) needs more work. I can think of a proof using von Dyck's theorem but it goes much beyond Galois theory. – user128787 Aug 10 '17 at 10:29
• @user128787 by sufficient, I mean: in general, suppose we have a group $G$ which is generated by $a$ and $b$, and that $|a| = n$, $|b| = m$, $ab = ba^k$, $k \ge 1$. Can't we say that $G = \langle a,b\mid a^n=b^m=1, ab=ba^k\rangle?$ – Cauchy Aug 10 '17 at 10:31
• Yes, as long as there is no other relation between $a$ and $b$. – user128787 Aug 10 '17 at 10:35
• @user128787 but could there be? If we know how to move $a$ from the left of $b$ to its right, don't we know everything else? – Cauchy Aug 10 '17 at 10:35
• @user128787 sorry if this is too naive of a question, but I still don't fully understand presentations. – Cauchy Aug 10 '17 at 10:36

Let $FG^{(2)}$ be the free group on $x,y$ and $K$ be the normal subgroup generated by $x^8,y^2,xyx^{-3}y^{-1}$, and let $G$ be as in the question. Consider the epimorphism $\phi: FG^{(2)} \to G$, $\phi(x) = \sigma$, $\phi(y) = \tau$. We have $K \subset \ker \phi$ so we have the induced epimorphism $\psi: FG^{(2)}/K \to G$, $\psi(a + K) = \phi(a)$, and $\left| FG^{(2)}/K \right| \ge |G| = 16$.
Now in $FG^{(2)}/K$, $y =y^{-1}$ and we have:
$$x^{-1} y = (yx)^{-1} = (xyx^{-2})^{-1} = x^2 y^{-1} x^{-1} = x^2 yx^{-1} =x(yx^3)x^{-1} = xyx^2 = yx^5$$
which shows that any element of $FG^{(2)}/K$ has the form $x^n$ or $yx^n$ for some $0\le n\le 7$. Since there are at most $16$ of those, we get $\left|FG^{(2)}/K\right| \le 16$ and therefore $G \cong FG^{(2)}/K$, so it indeed has the desired presentation.