Return to Answer

While Geoff Robinson has solved the problem for $p=2$, it may be worth to point out that the answer is "No" for all primes, for the following reason: It is known (see Curtis-Reiner, Methods of Rep'n Theory, §33) that the group ring $\mathbb{Z}_p G$ has infinite representation type, if $G$ is a cyclic $p$-group of order $\geq p^3$ (or if $G$ is a non-cyclic $p$-group, but that is not relevant here). On the other hand, $\mathbb{F}_p G$ has finite representation type (for $G$ cyclic). So we find an indecomposable $\mathbb{Z}_p G$-lattice $M$ that decomposes when reduced mod $p$. On the other hand, we may lift the summands of $M/pM$ to $\mathbb{Z}_p G$-lattices and form their direct sum, $N$ (say). Then $M/pM\cong N/pN$, but $M\not\cong N$.
EDIT: As has been poited out in the comments, the former is not correct. However, it follows from the proof of Dade's theorem given in Curtis-Reiner (33.8) that there are indecomposable, faithful $\mathbb{Z}_pG$-lattices of rank $k\left| G\right|$ for all $k$. As Alex Bartel points out in his answer, it follows from this fact that for some $k$ big enough, there must be non-isomorphic lattices of rank $k|G|$ reducing to isomorphic modules mod $p$. However, while I didn't check this, it seems to me that the indecomposable lattices of rank $k|G|$ constructed in the proof of Dade's theorem reduce to $(\mathbb{F}_pG)^k$ mod $p$, as does $(\mathbb{Z}_p G)^k $, of course. If correct, this gives concrete counterexamples, the smallest of dimension $2|G|$.
End EDIT.
More generally, the result is wrong if $p^3$ divides $m$. On the positive side, it is true when $p$ does not divide $m$. (Added later: This is elementary. Remember that $1+ p M_n(\mathbb{Z}_p) \subseteq GL_n(\mathbb{Z}_p)$, so units of $M_n(\mathbb{F}_p)$ lift to units of $M_n(\mathbb{Z}_p)$. After replacing $B$ with a conjugate, we may assume that $A\equiv B \mod p$. Then
$$ U:= \frac{1}{m} \sum_{k=0}^{m-1} A^{-k} B^k \equiv \frac{1}{m} \sum_{k=0}^{m-1} I \equiv I \mod p $$ which implies that $U$ is invertible. One computes that $AU=UB$, so it follows $A^U = B$.)

While Geoff Robinson has solved the problem for $p=2$, it may be worth to point out that the answer is "No" for all primes, for the following reason: It is known (see Curtis–Reiner, Methods of Rep'n Theory, §33) that the group ring $\mathbb{Z}_p G$ has infinite representation type, if $G$ is a cyclic $p$-group of order $\geq p^3$ (or if $G$ is a non-cyclic $p$-group, but that is not relevant here). On the other hand, $\mathbb{F}_p G$ has finite representation type (for $G$ cyclic). So we find an indecomposable $\mathbb{Z}_p G$-lattice $M$ that decomposes when reduced mod $p$. On the other hand, we may lift the summands of $M/pM$ to $\mathbb{Z}_p G$-lattices and form their direct sum, $N$ (say). Then $M/pM\cong N/pN$, but $M\not\cong N$.
EDIT: As has been pointed out in the comments, the former is not correct. However, it follows from the proof of Dade's theorem given in Curtis–Reiner (33.8) that there are indecomposable, faithful $\mathbb{Z}_pG$-lattices of rank $k\left| G\right|$ for all $k$. As Alex Bartel points out in his answer, it follows from this fact that for some $k$ big enough, there must be non-isomorphic lattices of rank $k\lvert G\rvert$ reducing to isomorphic modules mod $p$. However, while I didn't check this, it seems to me that the indecomposable lattices of rank $k\lvert G\rvert$ constructed in the proof of Dade's theorem reduce to $(\mathbb{F}_pG)^k$ mod $p$, as does $(\mathbb{Z}_p G)^k $, of course. If correct, this gives concrete counterexamples, the smallest of dimension $2\lvert G\rvert$.
End EDIT.
More generally, the result is wrong if $p^3$ divides $m$. On the positive side, it is true when $p$ does not divide $m$. (Added later: This is elementary. Remember that $1+ p M_n(\mathbb{Z}_p) \subseteq \operatorname{GL}_n(\mathbb{Z}_p)$, so units of $M_n(\mathbb{F}_p)$ lift to units of $M_n(\mathbb{Z}_p)$. After replacing $B$ with a conjugate, we may assume that $A\equiv B \mod p$. Then
$$ U:= \frac{1}{m} \sum_{k=0}^{m-1} A^{-k} B^k \equiv \frac{1}{m} \sum_{k=0}^{m-1} I \equiv I \mod p $$ which implies that $U$ is invertible. One computes that $AU=UB$, so it follows $A^U = B$.)

While Geoff Robinson has solved the problem for $p=2$, it may be worth to point out that the answer is "No" for all primes, for the following reason: It is known (see Curtis-Reiner, Methods of Rep'n Theory, §33) that the group ring $\mathbb{Z}_p G$ has infinite representation type, if $G$ is a cyclic $p$-group of order $\geq p^3$ (or if $G$ is a non-cyclic $p$-group, but that is not relevant here). On the other hand, $\mathbb{F}_p G$ has finite representation type (for $G$ cyclic). So we find an indecomposable $\mathbb{Z}_p G$-lattice $M$ that decomposes when reduced mod $p$. On the other hand, we may lift the summands of $M/pM$ to $\mathbb{Z}_p G$-lattices and form their direct sum, $N$ (say). Then $M/pM\cong N/pN$, but $M\not\cong N$.
EDIT: As has been poited out in the comments, the former is not correct. However, it follows from the proof of Dade's theorem given in Curtis-Reiner (33.8) that there are indecomposable, faithful $\mathbb{Z}_pG$-lattices of rank $k\left| G\right|$ for all $k$. As Alex Bartel points out in his answer, it follows from this fact that for some $k$ big enough, there must be non-isomorphic lattices of rank $k|G|$ reducing to isomorphic modules mod $p$. However, while I didn't check this, it seems to me that the indecomposable lattices of rank $k|G|$ constructed in the proof of Dade's theorem reduce to $(\mathbb{F}_pG)^k$ mod $p$, as does $(\mathbb{Z}_p G)^k$, of course. If correct, this gives concrete counterexamples, the smallest of dimension $2|G|$.
End EDIT.
More generally, the result is wrong if $p^3$ divides $m$. On the positive side, it is true when $p$ does not divide $m$.

While Geoff Robinson has solved the problem for $p=2$, it may be worth to point out that the answer is "No" for all primes, for the following reason: It is known (see Curtis-Reiner, Methods of Rep'n Theory, §33) that the group ring $\mathbb{Z}_p G$ has infinite representation type, if $G$ is a cyclic $p$-group of order $\geq p^3$ (or if $G$ is a non-cyclic $p$-group, but that is not relevant here). On the other hand, $\mathbb{F}_p G$ has finite representation type (for $G$ cyclic). So we find an indecomposable $\mathbb{Z}_p G$-lattice $M$ that decomposes when reduced mod $p$. On the other hand, we may lift the summands of $M/pM$ to $\mathbb{Z}_p G$-lattices and form their direct sum, $N$ (say). Then $M/pM\cong N/pN$, but $M\not\cong N$.
EDIT: As has been poited out in the comments, the former is not correct. However, it follows from the proof of Dade's theorem given in Curtis-Reiner (33.8) that there are indecomposable, faithful $\mathbb{Z}_pG$-lattices of rank $k\left| G\right|$ for all $k$. As Alex Bartel points out in his answer, it follows from this fact that for some $k$ big enough, there must be non-isomorphic lattices of rank $k|G|$ reducing to isomorphic modules mod $p$. However, while I didn't check this, it seems to me that the indecomposable lattices of rank $k|G|$ constructed in the proof of Dade's theorem reduce to $(\mathbb{F}_pG)^k$ mod $p$, as does $(\mathbb{Z}_p G)^k $, of course. If correct, this gives concrete counterexamples, the smallest of dimension $2|G|$.
End EDIT.
More generally, the result is wrong if $p^3$ divides $m$. On the positive side, it is true when $p$ does not divide $m$. (Added later: This is elementary. Remember that $1+ p M_n(\mathbb{Z}_p) \subseteq GL_n(\mathbb{Z}_p)$, so units of $M_n(\mathbb{F}_p)$ lift to units of $M_n(\mathbb{Z}_p)$. After replacing $B$ with a conjugate, we may assume that $A\equiv B \mod p$. Then
$$ U:= \frac{1}{m} \sum_{k=0}^{m-1} A^{-k} B^k \equiv \frac{1}{m} \sum_{k=0}^{m-1} I \equiv I \mod p $$ which implies that $U$ is invertible. One computes that $AU=UB$, so it follows $A^U = B$.)

While Geoff Robinson has solved the problem for $p=2$, it may be worth to point out that the answer is "No" for all primes, for the following reason: It is known (see Curtis-Reiner, Methods of Rep'n Theory, §33) that the group ring $\mathbb{Z}_p G$ has infinite representation type, if $G$ is a cyclic $p$-group of order $\geq p^3$ (or if $G$ is a non-cyclic $p$-group, but that is not relevant here). On the other hand, $\mathbb{F}_p G$ has finite representation type (for $G$ cyclic). So we find an indecomposable $\mathbb{Z}_p G$-lattice $M$ that decomposes when reduced mod $p$. On the other hand, we may lift the summands of $M/pM$ to $\mathbb{Z}_p G$-lattices and form their direct sum, $N$ (say). Then $M/pM\cong N/pN$, but $M\not\cong N$.
EDIT: As has been poited out in the comments, the former is not correct. However, it follows from the proof of Dade's theorem given in Curtis-Reiner (33.8) that there are indecomposable, faithful $\mathbb{Z}_pG$-lattices of rank $k\left| G\right|$ for all $k$. As Alex Bartel points out in his answer, it follows from this fact that for some $k$ big enough, there must be non-isomorphic lattices of rank $k|G|$ reducing to isomorphic lattices mod $p$. However, while I didn't check this, it seems to me that the indecomposable lattices of rank $k|G|$ constructed in the proof of Dade's theorem reduce to $(\mathbb{F}_pG)^k$ mod $p$, as does $(\mathbb{Z}_p G)^k$, of course. If correct, this gives concrete counterexamples, the smallest of dimension $2|G|$.
End EDIT.
More generally, the result is wrong if $p^3$ divides $m$. On the positive side, it is true when $p$ does not divide $m$.

While Geoff Robinson has solved the problem for $p=2$, it may be worth to point out that the answer is "No" for all primes, for the following reason: It is known (see Curtis-Reiner, Methods of Rep'n Theory, §33) that the group ring $\mathbb{Z}_p G$ has infinite representation type, if $G$ is a cyclic $p$-group of order $\geq p^3$ (or if $G$ is a non-cyclic $p$-group, but that is not relevant here). On the other hand, $\mathbb{F}_p G$ has finite representation type (for $G$ cyclic). So we find an indecomposable $\mathbb{Z}_p G$-lattice $M$ that decomposes when reduced mod $p$. On the other hand, we may lift the summands of $M/pM$ to $\mathbb{Z}_p G$-lattices and form their direct sum, $N$ (say). Then $M/pM\cong N/pN$, but $M\not\cong N$.
EDIT: As has been poited out in the comments, the former is not correct. However, it follows from the proof of Dade's theorem given in Curtis-Reiner (33.8) that there are indecomposable, faithful $\mathbb{Z}_pG$-lattices of rank $k\left| G\right|$ for all $k$. As Alex Bartel points out in his answer, it follows from this fact that for some $k$ big enough, there must be non-isomorphic lattices of rank $k|G|$ reducing to isomorphic modules mod $p$. However, while I didn't check this, it seems to me that the indecomposable lattices of rank $k|G|$ constructed in the proof of Dade's theorem reduce to $(\mathbb{F}_pG)^k$ mod $p$, as does $(\mathbb{Z}_p G)^k$, of course. If correct, this gives concrete counterexamples, the smallest of dimension $2|G|$.
End EDIT.
More generally, the result is wrong if $p^3$ divides $m$. On the positive side, it is true when $p$ does not divide $m$.