I am a beginner at group theory (and also a bit rusty at vectors) and have come across this question:
In the set of vectors with 3 components with the closed binary operation of vector product, find the identity element, if it exists, and if it does exist, find the inverse of a general element $a$, if it exists.
To find the element I know that $a \circ e = e \circ a = a$, where $e$ is the identity element. So if $a = \begin{pmatrix} x \\ y \\ z \end{pmatrix}$ and $e = \begin{pmatrix} i \\ j \\ k \end{pmatrix}$ then,
$$\begin{pmatrix} x \\ y \\ z \end{pmatrix} \times \begin{pmatrix} i \\ j \\ k \end{pmatrix} = \begin{pmatrix} i \\ j \\ k \end{pmatrix} \times \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} x \\ y \\ z \end{pmatrix} \\ \begin{pmatrix} yk-zj \\ zi-xk \\ xj-yi\end{pmatrix} = \begin{pmatrix} jz-ky \\ kx - iz \\ iy - jx\end{pmatrix} = \begin{pmatrix} x \\ y \\ z \end{pmatrix}$$ Looking at the top component of the vectors gives, $$\begin{align} yk-zj &= zj-yk \\ yk - zj &= -(yk-zj) \\ 1 &= -1\end{align}$$ And the same occurs for each other component. So does this mean that there is no identity element, or have I made a mistake somewhere?
Yes, there is no identity element for that operation. But you arrived too fast to the conclusion that $1=-1$; you can only divide by a number ($yk-zj$, in your case) when that number is not $0$.
Note that your're after a vector $e$ such that $a\times e=e\times a=a$. But $a\times e=-e\times a$. So, you can only have $a\times e=e\times a=a$ when $a=(0,0,0)$. So, there is no such element $e$.