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 $G$.
 A: 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 $\,p^n<p^{a_k}\,$...
