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Membership follows from $\mathbf{NL}=\mathbf{co}\text{-}\mathbf{NL}$. A directed graph is not strongly connected exactly when there is a pair $(u,v)$ for which $v$ is not reachable from $u$. Guessing the pair and using the NL procedure for non-reachability puts non-strong-connectivity in NL, hence strong connectivity is in co-NL and therefore in NL.

For NL-hardness, reduce directed reachability. Given $(G,s,t)$, form $G^{\prime }$ by retaining all edges of $G$, adding $v\to s$ for every vertex $v$, and adding $t\to v$ for every vertex $v$. If $t$ is reachable from $s$ in $G$, then any $u$ reaches any $v$ in $G^{\prime }$ alongConversely, if $G^{\prime }$ is strongly connected then $s$ reaches $t$. A simple $s$-to-$t$ path cannot use an added edge out of $t$ before arriving at $t$, and every added edge into $s$ merely returns the path to its starting vertex; deleting the resulting cycle leaves an $s$-to-$t$ path made from edges of $G$. The construction is computable in logarithmic space, so deciding strong connectivity is