# Showing two abelian groups have the same rank

Let $G=C_{p^{a_{1}}}\times C_{p^{a_{2}}}\times...\times C_{p^{a_{t}}}$ where $a_{1}\geq a_{2}\geq...\geq a_{t}$ and $H\subseteq G^{P^n}$ for some integer $n$. Please prove if $n>a_{k}$ for some $k\in\{1,...,t\}$ then, $G$ and $\frac{G}{H}$ have equal rank.

The rank $G$ is minimal number generators of $G$.

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Hint: $G/H$ is a direct product of $t$ groups. To see that these groups are non-trivial, assume that $C_{p^{a_k}} \subset H$ for some $k$. Why is this impossible? –  m_l May 16 '12 at 7:55
What does $H\subseteq G^{P^n}$ mean? I can guess that you meant a lower-case $p$ and that this is assumed to be a prime number, but how am I to form a quotient of $G$ by a subgroup of a high Cartesian power of $G$? Do you mean the subgroup of elements with $p^n$-torsion or something like that? Also "$n>a_k$ for some $k\in\{1,...,t\}$" would seem to mean just $n>a_t$. –  Marc van Leeuwen May 16 '12 at 9:09
Marc, I think he meant $$G^{p^n}:=<x^{p^n}\,;\,x\in G>=\{x^{p^n}\,;\,x\in G\}\,$$ , as G is an abelian group –  DonAntonio May 16 '12 at 10:21

Let us answer this question following m_$1$'s idea: if we put $\,C_{p^{a_i}}=\langle c_i\rangle\,$ , then $$G=\langle\,c_1\,,\,c_2\,,\ldots\,,c_t\,\rangle\Longrightarrow G/H=\langle\,c_1H\,,\,c_2H\,,\ldots\,,c_tH\,\rangle$$
So we see an element in $\,G\,$ as a vector with $\,t\,$ coordinates and coordinatewise group operation.
Now, suppose $$H\leq G^{p^n}:= \{\,x^{p^n}\;\;;\;\;x\in G\,\}\,\,,\,\text{with}\,\,n>a_i\,\,\text{for some}\,\,1\leq i\leq t$$ (the set $\,G^{p^n}\,$is a sbgp. because we're in an abelian group), then: $$\text{for some}\,\,1\leq k\leq t\,\,,\,\,c_kH=H\Longleftrightarrow c_k\in H\Longleftrightarrow C_{p^{a_k}}\leq H\leq G^{p^n}$$But this is impossible since
$(1)\,$ If $\,a_k\leq n\,$ , then we'd have that $\,c_k=(1,\ldots,1,c_k,1\ldots,1)\in G^{p^n}\,$, which is impossible as if $\,c_k=x^{p^n}\,\,\text{for some} \,x\in G\,$ above then $\,x\,$ has to be exactly of the same form: $\,x=(1,...,1,y,1,...,1)\,,\,y\,$ in the $\,k-$position, but then, putting $p^n=p^{a_k+h}=p^{a_k}p^h\,\,,\,h\geq 0\,$ , we'd get$$x^{p^n}=(1,...,1,\left(y^{p^{a_k}}\right)^{p^h},1,...,1)=(1,1,...,1,....,1)=1\in G$$
$(2)\,$ if $\,a_k>n\,$ then reasoning coordinatewise as above: on the $\,k-$th coordinate we won't get all the elements of $\,C_{p^{a_k}}\,$ as $\,p^n\,$ powers of elements in $\,G\,$ (lest the cyclic group $\,C_{p^{a_k}}\,$ of order $\,p^{a_k}\,$ has order $\,p^n<p^{a_k}\,$...