Finding a homomorphism for groups of same exponent Let $B$ be a finite abelian group of exponent dividing m (the smallest integer $r$ with $r/m$ and $b^r=e \;\forall b\in B$). Pick a $b\in B$. Then we can find a 
$c\in\mathbb{Z}/m\mathbb{Z}$ with $\operatorname{ord}(b)=\operatorname{ord}(c)$. Is there a homomomorphism $h:B\to\mathbb{Z}/m\mathbb{Z}$ with $h(b)=c$?
Tried some thing with the fundamental theorem for abelian groups, but couldn't construct an explicit homomorphism with this property.
 A: The answer is yes. We start with $B$ cyclic of prime power order, then prime power order in general, then the general case. We may clearly assume that $m$ is the exponent of $B$. 
cyclic prime power order
If $B$ is cyclic of prime power order then $B\cong\mathbb{Z}/m\mathbb{Z}$, and $m=p^t$ for some prime $p$ and integer $t$. Fix a generator $g$ of $B$. It suffices to do the following:

For any $x\in B$ of order $p^s$ find an automorphism mapping $g^{p^{t-s}}$ to $x$.

This is sufficient as we may map $b\mapsto g^{p^{t-s}}\mapsto c$ for the appropriate choice of $s$. 
Clearly any element can be written in the form $x=g^{x_0+x_1p+\cdots x_{t-1}p^{t-1}}$ with $x_i\in\{0,\ldots,p-1\}$ for each $i$ - we keep this notation, so for example $b=g^{b_0+b_1p+\cdots b_{t-1}p^{t-1}}$. For $x\ne 1$ let $l(x)$ be minimal such that $x_{l(x)}\ne 0$ - notably ${\rm ord}(x)=p^{t-l(x)}$. It will be convenient to choose $l(1)=t$.
For fixed $x$ consider the isomorphism $h:g\mapsto g^{x_{l(x)}+px_{l(x)+1}+\cdots+p^{t-l(x)-1}x_{t-1}}$ so $h(g^{p^{l(x)}})=x$ and we are done.
prime power order
Write $B=\mathbb{Z}/p^{e_1}\mathbb{Z}\times\cdots\times\mathbb{Z}/p^{e_t}\mathbb{Z}$ for some prime $p$ and integers $e_1\le\ldots\le e_t$ and let $f_i:B\to\mathbb{Z}/p^{e_i}\mathbb{Z}$ be the natural projection map for each $i$. Clearly $m=p^{e_t}$ and ${\rm ord}(b)=\max({\rm ord}(f_i(b)))$ so we may fix $s$ such that ${\rm ord}(b)={\rm ord}(f_s(b))$. We also fix $g_i\in B$ which generates $\mathbb{Z}/p^{e_i}\mathbb{Z}$ for each $i$. 
Let $h_0:\mathbb{Z}/p^{e_s}\mathbb{Z}\to\mathbb{Z}/p^{e_t}\mathbb{Z}$ be the injective homomorphism taking $\overline 1$ to $\overline{p^{e_t-e_s}}$. By the cyclic prime power order case there is some $h_1:\mathbb{Z}/p^{e_t}\mathbb{Z}\to\mathbb{Z}/p^{e_t}\mathbb{Z}$ with $h_1(h_0(f_s(b)))=c$. Now, let $h:B\to\mathbb{Z}/p^{e_t}\mathbb{Z}$ be the homomorphism with $h(g_i)=\overline{0}$ for $i\ne s$ and $h(g_s)=h_1(h_0(f_s(g_s)))$, then $h(b)=c$. 
general case
It is an immediate (depending on the precise statement) corollary of the fundamental theorem of finite abelian groups, that we may write $B=P_1\times\cdots\times P_t$ where the $P_i$ have pairwise coprime orders and each $P_i$ is of prime power order. Denote $b=(b_1,\ldots,b_t)$.
It is also straightforward to show that if $m_i$ is the exponent of $P_i$ for each $i$ then the $m_i$ are pairwise coprime and $m=m_1m_2\cdots m_t$. In particular $\mathbb{Z}/m\mathbb{Z}\cong\mathbb{Z}/m_1\mathbb{Z}\times\cdots\times\mathbb{Z}/m_t\mathbb{Z}$ so we may denote $c=(c_1,\ldots,c_t)$.
By the prime power order case, there is a homomorphism $h_i:P_i\to\mathbb{Z}/m_i\mathbb{Z}$ with $h_i(b_i)=c_i$ for each $i$, so the proof is complete by taking $h=h_1\times\cdots\times h_t$.
