Let $\cdot |_k\gamma$ denote the $k^{th}$ Petersson slash operator and let $j_\gamma(z) = cz + d$ where $\gamma = \begin{pmatrix} a &b\\ c& d\end{pmatrix} \in \operatorname{SL}_2(\Bbb Z)$ (so that $(f|_k\gamma )(z) = j_\gamma(z)^{-k}f(\gamma z)$).
Consider the series
$$\sum_{\gamma \in \operatorname{SL}_2(\Bbb Z)}(1 |_k\gamma)(z)$$
for all $z \in \frak h$, the upper half plane.
Obviously this is not a convergent series (since $\left(1|_k\begin{pmatrix}1 &b\\ 0 &1 \end{pmatrix}\right)(z) = 1$ for all $b \in \Bbb Z$ independent of $z$).
As a step towards improving on this, my lecture notes note that
$$(1_k\gamma)(z) = (1_k{\gamma^\prime})(z)\ \ \ \forall z \in \mathfrak{h} \iff \operatorname{SL}_2(\Bbb Z)_\infty\cdot \gamma = \operatorname{SL}_2(\Bbb Z)_\infty \cdot \gamma^\prime$$
where $\operatorname{SL}_2(\Bbb Z)_\infty$ is the stabiliser of $\infty$ in $\operatorname{SL}_2(\Bbb Z)$ and $\operatorname{SL}_2(\Bbb Z)_\infty$ acts on the left of $\operatorname{SL}_2(\Bbb Z)$. The notes then go on to say that the left side of this equivalence is "obviously" equivalent to
$$j_{\gamma\gamma^{\prime -1}}(z) = (1 |_k\gamma\gamma^{\prime - 1})(z) = 1.$$
Why is this obvious?
Hopefully this isn't just a silly oversight; naively multiplying out the automorphy factor and applying the cocycle condition doesn't seem to be helping me.
We have $\textrm{SL}_2(\mathbb{Z})_{\infty}\gamma = \textrm{SL}_2(\mathbb{Z})_{\infty}\gamma'$ if and only if $\gamma(\gamma')^{-1} \in \textrm{SL}_2(\mathbb{Z})_{\infty}$. Equivalently, $$\gamma(\gamma')^{-1} = \pm \begin{pmatrix} 1 & n \\ 0 & 1\end{pmatrix}$$ for some $n \in \mathbb{Z}$. This is the case if and only if the bottom row vector of $\gamma(\gamma')^{-1}$ is $(0,\pm 1)$, which is equivalent to $$j_{\gamma(\gamma')^{-1}}(z) = (\pm 1)^k.$$ Since you seem to be en route to constructing the Eisenstein series for $\textrm{SL}_2(\mathbb{Z})$, I am guessing you are assuming $k$ is even so that $(\pm 1)^k = 1$.