The question is from the proof of a theorem in Hardy's An Introduction to the Theory of Numbers.
THEOREM 57. If $(k, m) = d$, then the congruence $(5.4.1)\qquad kx \equiv l \pmod{m}$ is soluble if and only if $d|l$. It has then just $d$ solutions. In particular, if $(k, m) = 1$, the congruence has always just one solution.
Proof:
The congruence is equivalent to $kx-my=l,$
so that the result is partly contained in Theorem 25. It is naturally to be understood, when we say that the congruence has "just d" solutions, that congruent solutions are regarded as the same.
If $d = 1$, then Theorem 57 is a corollary of Theorem 56. If $d > 1$, the congruence (5.4.1) is clearly insoluble unless $d|l$. If $d|l$, then $m = dm', \quad k = dk',\quad l = dl',$ and the congruence is equivalent to $(5.4.2)\qquad k'x \equiv l'\pmod{m'}.$ Since $(k', m') = 1$, (5.4.2) has just one solution.
If this solution is $x\equiv t\pmod{ m'},$ then $x=t+ym',$ and the complete set of solutions of (5.4.1) is found by giving $y$ all values which lead to values of $t + ym'$ incongruent to modulus $m$. Since $t+ym'\equiv t+zm'\pmod{m}\Leftrightarrow m|m'(y-z)\Leftrightarrow d|(y-z),$ there are just $d$ solutions, represented by $t, t + m', t + 2m',\cdots, t+(d-1)m'.$ This proves the theorem.
Originally, the second last sentence is
Here are my questions:
Is there a typo in the original text?
Why are the $d$ solutions represented like that? What do the commas mean here? (Do they mean "or" or "and"?)
What's the relationship between the last step in the proof and the question here?