Groups in a category

Question

Steve Awodey defines a group in a category as a diagram (group) satisfying the conditions that the following diagrams commute: first, second, third. But then he says:

Note that the requirement that these diagrams commute is equivalent to the more familiar condition that, for all (generalized) elements, $x,y,z: Z \rightarrow{} G$ the following equations hold:$m(m(x, y), z) = m(x,m(y, z))\\ m(x, u) = x = m(u, x)\\ m(x, ix) = u = m(ix, x)$

How do these equations follow from the definition of a group? Do i understand correctly that $m(x,y)$ stands for $m \circ \left \langle x,y \right \rangle$? And if it is, how to derive from $m \circ (m\times 1) = m \circ (1 \times m) \circ \cong$ (that corresponds to the first diagram) this equation: $m \circ \left \langle m \circ\left \langle x,y \right \rangle,z \right \rangle = m \circ \left \langle x,m\circ\left \langle y,z \right \rangle \right \rangle$ for all (generalized) elements $x,y,z:Z\rightarrow G$

Answer 1

Just pre-compose both sides of $m\circ(m\times id_G)=m\circ(id_G\times m)\circ\alpha$ with $\langle\langle x,y\rangle, z\rangle$, $\alpha$ being the associator, and start computing. (You didn't ask for it, but for the other direction just pick $Z=G$ and $x$, $y$, and $z$ as as the appropriate projections, specifically $\pi_1\circ\pi_1$, $\pi_2\circ\pi_1$, and $\pi_2$ respectively.)

Answer 2

One has to realise that $m:(x,y)\to m(x,y)$ is the multiplication of the group, $u$ the identity and $ix$ the inverse of $x$ and we have the more familiar definition of a group.

So we have just rewritten in arrows associativity, the definition of identity and the definition of an inverse element