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Robin Chapman
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For a group $G=\mathbb{Z}_{p^r}^k$ this is quite straightforward. Let $g$ have order $p^r$ in $G$ (if not then we are effectively working in $\mathbb{Z}_{p^s}$ where $s < k$). Applying an automorphism of $G$ we can assume that $g=(1,0,\ldots,0)$. Endomorphisms of $G$ correspond to matrices over $\mathbb{Z}_{p^r}$. The image of $s$ is the first row (or column if you put the map on the other side) and we see that the image of $s$ is uniformly distibuted over all of $G$.

Now consider a general finite abelian $p$-group $G$. Let $g\in G$. We can write $G=\langle h\rangle\times H$ where $g=p^s h\in H$ and $h$ has order $p^m$ for some $m\ge s$. We can specify an endomorphism of $G$ by mapping $h$ to any element $h'$ of order $\le p^m$ and taking any homomorphism from $H$ to $G$. Then the $h'$ are uniformly distributed amongst the elements of order $\le p^m$ in $G$ and $g'$ is mapped to $g'=p^{m-r}h'$. These $g'$ are uniformly distributed over a certain subgroup of $G$.

For general finite abelian $G$ split up $G$ as a product of its Sylow $p$-subgroups. Then $g\in G$ splits up into its primary components and each of these behave in the same way, under a random endomorphism, as in the $p$-group case above.

Added I now see that the argument I gave in the prime power case is valid in the general case too. The key observation is that a maximal cyclic subgroup of a finite abelian group is a direct summand. Let $g\in G$ have order $m$ and let $H$ be a maximal cyclic subgroup of order $mn$ containing $\langle g\rangle$. Then the images of $g$ under random endomorphisms of $G$ are uniformly distributed in the subgroup $n G[mn]$ of $G$ where $G[mn]$ denotes the $mn$-torsion subgroup of $G$.

Added (4/7/2010) Thanks to Tom for pointing out my error above. The argument I had in mind for proving that maximal cyclic groups are summands doesn't actually work. :-( As t3suji points out, the images are uniformly distributed over a subgroup. Identifying this subgroup looks like being a bit more fiddly than I believed and I lack the patience to do it now. It seems that reduction to the prime power case is a good way to proceed.

For a group $G=\mathbb{Z}_{p^r}^k$ this is quite straightforward. Let $g$ have order $p^r$ in $G$ (if not then we are effectively working in $\mathbb{Z}_{p^s}$ where $s < k$). Applying an automorphism of $G$ we can assume that $g=(1,0,\ldots,0)$. Endomorphisms of $G$ correspond to matrices over $\mathbb{Z}_{p^r}$. The image of $s$ is the first row (or column if you put the map on the other side) and we see that the image of $s$ is uniformly distibuted over all of $G$.

Now consider a general finite abelian $p$-group $G$. Let $g\in G$. We can write $G=\langle h\rangle\times H$ where $g=p^s h\in H$ and $h$ has order $p^m$ for some $m\ge s$. We can specify an endomorphism of $G$ by mapping $h$ to any element $h'$ of order $\le p^m$ and taking any homomorphism from $H$ to $G$. Then the $h'$ are uniformly distributed amongst the elements of order $\le p^m$ in $G$ and $g'$ is mapped to $g'=p^{m-r}h'$. These $g'$ are uniformly distributed over a certain subgroup of $G$.

For general finite abelian $G$ split up $G$ as a product of its Sylow $p$-subgroups. Then $g\in G$ splits up into its primary components and each of these behave in the same way, under a random endomorphism, as in the $p$-group case above.

Added I now see that the argument I gave in the prime power case is valid in the general case too. The key observation is that a maximal cyclic subgroup of a finite abelian group is a direct summand. Let $g\in G$ have order $m$ and let $H$ be a maximal cyclic subgroup of order $mn$ containing $\langle g\rangle$. Then the images of $g$ under random endomorphisms of $G$ are uniformly distributed in the subgroup $n G[mn]$ of $G$ where $G[mn]$ denotes the $mn$-torsion subgroup of $G$.

For a group $G=\mathbb{Z}_{p^r}^k$ this is quite straightforward. Let $g$ have order $p^r$ in $G$ (if not then we are effectively working in $\mathbb{Z}_{p^s}$ where $s < k$). Applying an automorphism of $G$ we can assume that $g=(1,0,\ldots,0)$. Endomorphisms of $G$ correspond to matrices over $\mathbb{Z}_{p^r}$. The image of $s$ is the first row (or column if you put the map on the other side) and we see that the image of $s$ is uniformly distibuted over all of $G$.

Now consider a general finite abelian $p$-group $G$. Let $g\in G$. We can write $G=\langle h\rangle\times H$ where $g=p^s h\in H$ and $h$ has order $p^m$ for some $m\ge s$. We can specify an endomorphism of $G$ by mapping $h$ to any element $h'$ of order $\le p^m$ and taking any homomorphism from $H$ to $G$. Then the $h'$ are uniformly distributed amongst the elements of order $\le p^m$ in $G$ and $g'$ is mapped to $g'=p^{m-r}h'$. These $g'$ are uniformly distributed over a certain subgroup of $G$.

For general finite abelian $G$ split up $G$ as a product of its Sylow $p$-subgroups. Then $g\in G$ splits up into its primary components and each of these behave in the same way, under a random endomorphism, as in the $p$-group case above.

Added I now see that the argument I gave in the prime power case is valid in the general case too. The key observation is that a maximal cyclic subgroup of a finite abelian group is a direct summand. Let $g\in G$ have order $m$ and let $H$ be a maximal cyclic subgroup of order $mn$ containing $\langle g\rangle$. Then the images of $g$ under random endomorphisms of $G$ are uniformly distributed in the subgroup $n G[mn]$ of $G$ where $G[mn]$ denotes the $mn$-torsion subgroup of $G$.

Added (4/7/2010) Thanks to Tom for pointing out my error above. The argument I had in mind for proving that maximal cyclic groups are summands doesn't actually work. :-( As t3suji points out, the images are uniformly distributed over a subgroup. Identifying this subgroup looks like being a bit more fiddly than I believed and I lack the patience to do it now. It seems that reduction to the prime power case is a good way to proceed.

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Robin Chapman
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For a group $G=\mathbb{Z}_{p^r}^k$ this is quite straightforward. Let $g$ have order $p^r$ in $G$ (if not then we are effectively working in $\mathbb{Z}_{p^s}$ where $s < k$). Applying an automorphism of $G$ we can assume that $g=(1,0,\ldots,0)$. Endomorphisms of $G$ correspond to matrices over $\mathbb{Z}_{p^r}$. The image of $s$ is the first row (or column if you put the map on the other side) and we see that the image of $s$ is uniformly distibuted over all of $G$.

Now consider a general finite abelian $p$-group $G$. Let $g\in G$. We can write $G=\langle h\rangle\times H$ where $g=p^s h\in H$ and $h$ has order $p^m$ for some $m\ge s$. We can specify an endomorphism of $G$ by mapping $h$ to any element $h'$ of order $\le p^m$ and taking any homomorphism from $H$ to $G$. Then the $h'$ are uniformly distributed amongst the elements of order $\le p^m$ in $G$ and $g'$ is mapped to $g'=p^{m-r}h'$. These $g'$ are uniformly distributed over a certain subgroup of $G$.

For general finite abelian $G$ split up $G$ as a product of its Sylow $p$-subgroups. Then $g\in G$ splits up into its primary components and each of these behave in the same way, under a random endomorphism, as in the $p$-group case above.

Added I now see that the argument I gave in the prime power case is valid in the general case too. The key observation is that a maximal cyclic subgroup of a finite abelian group is a direct summand. Let $g\in G$ have order $m$ and let $H$ be a maximal cyclic subgroup of order $mn$ containing $\langle g\rangle$. Then the images of $g$ under random endomorphisms of $G$ are uniformly distributed in the subgroup $n G[mn]$ of $G$ where $G[mn]$ denotes the $mn$-torsion subgroup of $G$.

For a group $G=\mathbb{Z}_{p^r}^k$ this is quite straightforward. Let $g$ have order $p^r$ in $G$ (if not then we are effectively working in $\mathbb{Z}_{p^s}$ where $s < k$). Applying an automorphism of $G$ we can assume that $g=(1,0,\ldots,0)$. Endomorphisms of $G$ correspond to matrices over $\mathbb{Z}_{p^r}$. The image of $s$ is the first row (or column if you put the map on the other side) and we see that the image of $s$ is uniformly distibuted over all of $G$.

Now consider a general finite abelian $p$-group $G$. Let $g\in G$. We can write $G=\langle h\rangle\times H$ where $g=p^s h\in H$ and $h$ has order $p^m$ for some $m\ge s$. We can specify an endomorphism of $G$ by mapping $h$ to any element $h'$ of order $\le p^m$ and taking any homomorphism from $H$ to $G$. Then the $h'$ are uniformly distributed amongst the elements of order $\le p^m$ in $G$ and $g'$ is mapped to $g'=p^{m-r}h'$. These $g'$ are uniformly distributed over a certain subgroup of $G$.

For general finite abelian $G$ split up $G$ as a product of its Sylow $p$-subgroups. Then $g\in G$ splits up into its primary components and each of these behave in the same way, under a random endomorphism, as in the $p$-group case above.

For a group $G=\mathbb{Z}_{p^r}^k$ this is quite straightforward. Let $g$ have order $p^r$ in $G$ (if not then we are effectively working in $\mathbb{Z}_{p^s}$ where $s < k$). Applying an automorphism of $G$ we can assume that $g=(1,0,\ldots,0)$. Endomorphisms of $G$ correspond to matrices over $\mathbb{Z}_{p^r}$. The image of $s$ is the first row (or column if you put the map on the other side) and we see that the image of $s$ is uniformly distibuted over all of $G$.

Now consider a general finite abelian $p$-group $G$. Let $g\in G$. We can write $G=\langle h\rangle\times H$ where $g=p^s h\in H$ and $h$ has order $p^m$ for some $m\ge s$. We can specify an endomorphism of $G$ by mapping $h$ to any element $h'$ of order $\le p^m$ and taking any homomorphism from $H$ to $G$. Then the $h'$ are uniformly distributed amongst the elements of order $\le p^m$ in $G$ and $g'$ is mapped to $g'=p^{m-r}h'$. These $g'$ are uniformly distributed over a certain subgroup of $G$.

For general finite abelian $G$ split up $G$ as a product of its Sylow $p$-subgroups. Then $g\in G$ splits up into its primary components and each of these behave in the same way, under a random endomorphism, as in the $p$-group case above.

Added I now see that the argument I gave in the prime power case is valid in the general case too. The key observation is that a maximal cyclic subgroup of a finite abelian group is a direct summand. Let $g\in G$ have order $m$ and let $H$ be a maximal cyclic subgroup of order $mn$ containing $\langle g\rangle$. Then the images of $g$ under random endomorphisms of $G$ are uniformly distributed in the subgroup $n G[mn]$ of $G$ where $G[mn]$ denotes the $mn$-torsion subgroup of $G$.

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Robin Chapman
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For a group $G=\mathbb{Z}_{p^r}^k$ this is quite straightforward. Let $g$ have order $p^r$ in $G$ (if not then we are effectively working in $\mathbb{Z}_{p^s}$ where $s < k$). Applying an automorphism of $G$ we can assume that $g=(1,0,\ldots,0)$. Endomorphisms of $G$ correspond to matrices over $\mathbb{Z}_{p^r}$. The image of $s$ is the first row (or column if you put the map on the other side) and we see that the image of $s$ is uniformly distibuted over all of $G$.

Now consider a general finite abelian $p$-group $G$. There is a surjective homomorphismLet $\pi:H\to G$ where$g\in G$. We can write $H=\mathbb{Z}_{p^r}^k$$G=\langle h\rangle\times H$ for somewhere $r$$g=p^s h\in H$ and $k$$h$ has order $p^m$ for some $m\ge s$. EachWe can specify an endomorphism of $G$ lifts to an endomorphismby mapping of$h$ to any element $H$ and the number$h'$ of these lifted endomorphisms is the sameorder $\le p^m$ and taking any homomorphism for each endomorphism offrom $H$ to $G$. ThusThen the images of $g\in G$ under an arbitrary endomorphism of $G$$h'$ are the images under $\pi$uniformly distributed ofamongst the imageselements of a liftingorder $h$ of$\le p^m$ in $h$ under an arbitrary endomorphism of$G$ and $g'$ is $H$mapped to $g'=p^{m-r}h'$. Morevover these imagesThese $g'$ are uniformly distibuted over alldistributed admissible elements. The upshot is that the random endomorphismover a certain subgroup of $G$ takes $g$ to $\pi(h')$ where $\pi(h)=g$ and $h'\in H$ has the same order as $h$ and all such images are acheived with equal probability.

For general finite abelian $G$ split up $G$ as a product of its Sylow $p$-subgroups. Then $g\in G$ splits up into its primary components and each of these behave in the same way, under a random endomorphism, as in the $p$-group case above.

For a group $G=\mathbb{Z}_{p^r}^k$ this is quite straightforward. Let $g$ have order $p^r$ in $G$ (if not then we are effectively working in $\mathbb{Z}_{p^s}$ where $s < k$). Applying an automorphism of $G$ we can assume that $g=(1,0,\ldots,0)$. Endomorphisms of $G$ correspond to matrices over $\mathbb{Z}_{p^r}$. The image of $s$ is the first row (or column if you put the map on the other side) and we see that the image of $s$ is uniformly distibuted over all of $G$.

Now consider a general finite abelian $p$-group $G$. There is a surjective homomorphism $\pi:H\to G$ where $H=\mathbb{Z}_{p^r}^k$ for some $r$ and $k$. Each endomorphism of $G$ lifts to an endomorphism of $H$ and the number of these lifted endomorphisms is the same for each endomorphism of $G$. Thus the images of $g\in G$ under an arbitrary endomorphism of $G$ are the images under $\pi$ of the images of a lifting $h$ of $h$ under an arbitrary endomorphism of $H$. Morevover these images are uniformly distibuted over all admissible elements. The upshot is that the random endomorphism of $G$ takes $g$ to $\pi(h')$ where $\pi(h)=g$ and $h'\in H$ has the same order as $h$ and all such images are acheived with equal probability.

For general finite abelian $G$ split up $G$ as a product of its Sylow $p$-subgroups. Then $g\in G$ splits up into its primary components and each of these behave in the same way, under a random endomorphism, as in the $p$-group case above.

For a group $G=\mathbb{Z}_{p^r}^k$ this is quite straightforward. Let $g$ have order $p^r$ in $G$ (if not then we are effectively working in $\mathbb{Z}_{p^s}$ where $s < k$). Applying an automorphism of $G$ we can assume that $g=(1,0,\ldots,0)$. Endomorphisms of $G$ correspond to matrices over $\mathbb{Z}_{p^r}$. The image of $s$ is the first row (or column if you put the map on the other side) and we see that the image of $s$ is uniformly distibuted over all of $G$.

Now consider a general finite abelian $p$-group $G$. Let $g\in G$. We can write $G=\langle h\rangle\times H$ where $g=p^s h\in H$ and $h$ has order $p^m$ for some $m\ge s$. We can specify an endomorphism of $G$ by mapping $h$ to any element $h'$ of order $\le p^m$ and taking any homomorphism from $H$ to $G$. Then the $h'$ are uniformly distributed amongst the elements of order $\le p^m$ in $G$ and $g'$ is mapped to $g'=p^{m-r}h'$. These $g'$ are uniformly distributed over a certain subgroup of $G$.

For general finite abelian $G$ split up $G$ as a product of its Sylow $p$-subgroups. Then $g\in G$ splits up into its primary components and each of these behave in the same way, under a random endomorphism, as in the $p$-group case above.

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Robin Chapman
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