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To expand on Steve D's comment, the answer is indeed yes for finite abelian groups.

Proof. We proceed by (strong) induction on $|G|$. Let $G$ and $H$ be finite abelian groups for which there exists a bijection $f:G \to H$ that preserves subgroups.

Suppose $G=A \oplus B$, where $A$ and $B$ are both proper subgroups of $G$. By induction, $A \cong f(A)$ and $B \cong f(B)$. Since $A \cap B=\{0\}$, we conclude that $f(A) \cap f(B)=\{f(0)\}$. Since $f$ preserves subgroups, $f(A)$ and $f(B)$ are both subgroups of $H$, whose interection is the zero element of $H$. Moreover, since $|f(A)||f(B)|= |H|$, it follows that $H = f(A) \oplus f(B) \cong A \oplus B \cong G$. Therefore, we are done unless $G=\mathbb Z / p^k \mathbb Z$. However, by interchanging the roles of $G$ and $H$ we conclude that $H$ is also equal to $\mathbb Z / p^k \mathbb Z$.

To expand on Steve D's comment, the answer is indeed yes for finite abelian groups. The following is a simplified version of an earlier proof (rendering some of the below comments obsolete).

Proof. We proceed by (strong) induction on $|G|$. Let $G$ and $H$ be finite abelian groups for which there exists a bijection $f:G \to H$ that preserves subgroups.

Suppose $G=A \oplus B$, where $A$ and $B$ are both proper subgroups of $G$. By induction, $A \cong f(A)$ and $B \cong f(B)$. Since $A \cap B=\{0\}$, we conclude that $f(A) \cap f(B)=\{f(0)\}$. Since $f$ preserves subgroups, $f(A)$ and $f(B)$ are both subgroups of $H$, whose interection is the zero element of $H$. Moreover, since $|f(A)||f(B)|= |H|$, it follows that $H = f(A) \oplus f(B) \cong A \oplus B \cong G$. Therefore, we are done unless $G=\mathbb Z / p^k \mathbb Z$. However, by interchanging the roles of $G$ and $H$ we conclude that $H$ is also equal to $\mathbb Z / p^k \mathbb Z$.

To expand on Steve D's comment, I also believe the answer is yes for finite abelian groups. We can prove this by (strong) induction on $|G|$. Let $G$ and $H$ be finite abelian groups for which there exists a bijection $f:G \to H$ that preserves subgroups.

Suppose $G=A \oplus B$, where $A$ and $B$ are both proper subgroups of $G$. By induction, $A \cong f(A)$ and $B \cong f(B)$. Since $A \cap B=\{0\}$, we conclude that $f(A) \cap f(B)=\{f(0)\}$. Since $f$ preserves subgroups, $f(A)$ and $f(B)$ are both subgroups of $H$, whose interection is the zero element of $H$. Thus, $H = f(A) \oplus f(B) \cong A \oplus B \cong G$. Therefore, we are done unless $G=\mathbb Z / p^k \mathbb Z$. However, by interchanging the roles of $G$ and $H$ we conclude that $H$ is also equal to $\mathbb Z / p^k \mathbb Z$.

To expand on Steve D's comment, the answer is indeed yes for finite abelian groups.

Proof. We proceed by (strong) induction on $|G|$. Let $G$ and $H$ be finite abelian groups for which there exists a bijection $f:G \to H$ that preserves subgroups.

Suppose $G=A \oplus B$, where $A$ and $B$ are both proper subgroups of $G$. By induction, $A \cong f(A)$ and $B \cong f(B)$. Since $A \cap B=\{0\}$, we conclude that $f(A) \cap f(B)=\{f(0)\}$. Since $f$ preserves subgroups, $f(A)$ and $f(B)$ are both subgroups of $H$, whose interection is the zero element of $H$. Moreover, since $|f(A)||f(B)|= |H|$, it follows that $H = f(A) \oplus f(B) \cong A \oplus B \cong G$. Therefore, we are done unless $G=\mathbb Z / p^k \mathbb Z$. However, by interchanging the roles of $G$ and $H$ we conclude that $H$ is also equal to $\mathbb Z / p^k \mathbb Z$.

To expand on Steve D's comment, I also believe the answer is yes for finite abelian groups. We can prove this by (strong) induction on $|G|$. Let $G$ and $H$ be finite abelian groups for which there exists a bijection $f:G \to H$ that preserves subgroups.

First consider the case that both $G$ and $H$ are not cyclic. Let $A$ be a largest order cyclic subgroup of $G$. Observe that $G=A \times B$. By induction, $A \cong f(A)$, and $B \cong f(B)$. Moreover, by reversing the roles of $G$ and $H$, we may conclude that $f(A)$ is a largest order cyclic subgroup of $H$. Thus, $H \cong f(A) \times f(B) \cong A \times B \cong G$.

We may now assume that $G$ is cyclic. As above, we are done by induction unless $|G|$ is a prime power, say $|G|=p^m$. If $H$ is cyclic we are done. However, if $H$ is not cyclic, then it is easy to check that it has more subgroups than $G$. Thus, such a bijection $f$ cannot exist.

To expand on Steve D's comment, I also believe the answer is yes for finite abelian groups. We can prove this by (strong) induction on $|G|$. Let $G$ and $H$ be finite abelian groups for which there exists a bijection $f:G \to H$ that preserves subgroups.

Suppose $G=A \oplus B$, where $A$ and $B$ are both proper subgroups of $G$. By induction, $A \cong f(A)$ and $B \cong f(B)$. Since $A \cap B=\{0\}$, we conclude that $f(A) \cap f(B)=\{f(0)\}$. Since $f$ preserves subgroups, $f(A)$ and $f(B)$ are both subgroups of $H$, whose interection is the zero element of $H$. Thus, $H = f(A) \oplus f(B) \cong A \oplus B \cong G$. Therefore, we are done unless $G=\mathbb Z / p^k \mathbb Z$. However, by interchanging the roles of $G$ and $H$ we conclude that $H$ is also equal to $\mathbb Z / p^k \mathbb Z$.