Let $\mathcal{C}$ be a full subcategory of $\text{Mod}_A$, the category of modules over a commutative unital ring $A$ (I am mostly interested in the graded setting, but I doubt there is much difference here between graded and ungraded cases).
Let $I$ be an injective $A$-module contained in $\mathcal{C}$. Must $I$ be an injective object in $\mathcal{C}$? Let $M$ and $N$ be modules in $\mathcal{C}$, $\iota:M\to N$ be a monomorphism in $\mathcal{C}$, and $\alpha:M\to I$ be any morphism. If $\iota$ is also a monomorphism in $\text{Mod}_A$, then this gives a morphism $\beta:N\to I$ in $\mathcal{C}$ with $\beta\iota=\alpha$, so that $I$ is an injective object in $\mathcal{C}$.
But how do we know that $\iota$ is a monomorphism in $\text{Mod}_A$? Let $X$ be any $A$-module and $\varphi,\psi:X\to M$ with $\iota\varphi=\iota\psi$. Need to show that $\varphi=\psi$. But we only know from the fact that $\iota$ is mono in $\mathcal{C}$ that this works for $X\in\mathcal{C}$. Am I missing something?
EDIT: In the case I am most interested in, we have the following property: if $M, N\in\mathcal{C}$ and $f:M\to N$, then $\text{ker}(f)\in\mathcal{C}$. So in the argument above, $\text{ker}(\iota)\in\mathcal{C}$. Let $\varphi$ be the inclusion $\text{ker}(\iota)\to M$ and $\psi$ be zero, so that $\iota\varphi=\iota\psi$. Since $\iota$ is mono in $\mathcal{C}$, we know that $\varphi=\psi$, completing the proof.
Best Answer
The following is a counterexample: take $A = \mathbb{Z}$ and $\mathcal{C}$ be the full subcategory containing $\mathbb{Q}$ and $\mathbb{Q}/\mathbb{Z}$. Then I claim the natural quotient map $q: \mathbb{Q} \rightarrow \mathbb{Q}/\mathbb{Z}$ is a monomorphism. Granted the claim, it is clear then $\mathbb{Q}$ cannot be injective in $\mathcal{C}$ despite being an injective abelian group, as the identity morphism $\mathbb{Q} \rightarrow \mathbb{Q}$ cannot be extended to $\mathbb{Q}/\mathbb{Z}$.
Now, to prove the claim, we note that the only morphism from $\mathbb{Q}/\mathbb{Z}$ to $\mathbb{Q}$ is the zero map, so there’s nothing to verify there. Thus, it suffices to show that, if $f, g: \mathbb{Q} \rightarrow \mathbb{Q}$ are morphisms s.t. $q \circ f = q \circ g$, then $f = g$. We observe that $q \circ f = q \circ g$ means $q \circ (f - g) = 0$, whence $f - g$ is a morphism from $\mathbb{Q}$ to $\mathbb{Z}$. But the only such morphism is the zero map, so $f = g$.