Show that $\forall a\in\mathbb{H}, \ \exists b \in\mathbb{H}: ab =ba = 1.$
I am pretty sure I can easily google the multiplicative inverse in $\mathbb{H}$, but can you give me a hint on how to determine the inverse von an arbitrary $a \in\mathbb{h}$ myself? This task comes directly after showing that the conjugation is a ring antihomomorphism. ($a\ \bar+ \ b\ = \bar a+\bar b, a\ \bar *\ b = \bar b*\bar a$ and $ 1 = \bar1$)
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You should just solve for ($(x + y i + z j + w k)(a+b i + c j + d k) = 1.$ Notice that you will get a system of linear equations in $x, y, z, w.$
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This is a somewhat nonstandard approach, but I think it illustrates a nice model for $\mathbb H$.
Note that $\mathbb H$ is isomorphic to the algebra of matrices over $\mathbb C$ of the form $$\begin{pmatrix} a+bi & c+di\\ -c+di & a-bi \end{pmatrix}$$ which is easy to verify: you just need to check that it contains a copy of $\mathbb R$ (namely the matrices with $b=c=d=0$) and that it contains $i,j,k$ such that $i^2=j^2=k^2=ijk=-1$, which come from setting everything but $b,c$ and $d$ respectively equal to $0$.
Now the inverse is just the inverse of a $2\times 2$ matrix, which is easy to compute and can easily be shown to have the same form.
Inverses are easily found inside the quaternions with conjugation, multiplication and division.
Really, the only computational fact you need is that $a\overline{a}\in \Bbb R$. By looking at what conjugation does to quaternions, you can easily see that $\overline{q}=q\iff q\in \Bbb R$.
But then since conjugation is an antihomomorphism, $\overline{q\overline{q}}=\overline{\bar q}\bar q=q\bar q$, so $q\bar q$ is real.
But then look: $q(\frac{\bar q}{q\bar q})=1$ finds a right inverse for $q$. Since this is true for all nonzero quaternions, $\frac{\bar q}{q\bar q}$ has a right inverse also, which is necessarily $q$. Thus $\frac{\bar q}{q\bar q}=q^{-1}$.