The word problem for groups was shown unsolvable by (computably) associating to each Turing machine $M$ a finitely presented group $G_{M}$ such that $M$ halts on input word $w$ if and only if $w$ is equivalent to the identity in $G_{M}$. More precisely, let $A_{\text{TM}}$ be the accepting language for Turing machines (also known as the halting set) and let $\text{Ident}$ be the language of words $\langle{G,w}\rangle$ such that $w$ is equivalent to the identity in the group $G$. A computable function $f: A_{\text{TM}} \to \text{Ident}$ was constructed so that $\langle{M,w}\rangle \in A_{\text{TM}}$ $\Longleftrightarrow$ $f(\langle{M, w}\rangle) \in \text{Ident}$. Thus to decide if machine $M$ accepts input $w$ (i.e. if $\langle{M,w}\rangle \in A_{\text{TM}}$), it suffices to compute $f(\langle{M,w}\rangle)$ and decide if that is in $\text{Ident}$. But $A_{\text{TM}}$ is undecidable, so $\text{Ident}$ must also be.
However, this argument wouldn't work if we only had $\langle{M,w}\rangle \in A_{\text{TM}}$ $\Longrightarrow$ $f(\langle{M, w}\rangle) \in \text{Ident}$, because then, as Gadi and Carl pointed out, even if we could decide if $f(\langle{M,w}\rangle) \in \text{Ident}$, it would give us no information about whether or not $\langle{M,w}\rangle \in A_{\text{TM}}$ because in the case where $f(\langle{M,w}\rangle) \in \text{Ident}$ holds, we could still have $\langle{M,w}\rangle \notin A_{\text{TM}}$.
It is worth noting that, conversely, if we only had $\langle{M,w}\rangle \in A_{\text{TM}}$ $\Longleftarrow$ $f(\langle{M, w}\rangle) = \langle{G, w}\rangle \in \text{Ident}$, the case where $f(\langle{M,w}\rangle) \notin \text{Ident}$ gives no information.