I have two Abstract Algebra Problems 
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*I have a conjecture and I can't seem to find a counterexample, but I think it is false: Let $G$ be an abelian group. If $\{a_1, a_2, \dots , a_n\}$ is a minimal generating set for $G$ and for each $a_k$, $|a_k| = d_k \in \mathbb{Z}$, then each $g \in G$ has a unique representation as $g = a_1^{s_1}a_2^{s_2}...a_n^{s_n}$ where $0 \leq s_k < d_k$ for each $k$.


I know the fundamental theorem of finite abelian groups.


*I am trying to find a nontrivial homomorphism mapping from $D_4 = \langle a,b | a^4 = b^2 = abab = e\rangle$ to $\mathbb{Z}_8$. I am having trouble with this because $D_4$ is not abelian and I don't know how to relate its structure to $\mathbb{Z}_8$.

 A: 1) There are two possible interpretations of “minimal generating set”. Unfortunately, your conjecture is false under either one.
Interpretation 1. $\{a_1,\ldots,a_n\}$ is a minimal generating set for $G$ if $G=\langle a_1,\ldots,a_n\rangle$, but for any proper subset $X\subsetneq \{a_1,\ldots,a_n\}$, $G\neq\langle X\rangle$.
That is, the set generates, but no proper subset does.
For a counterexample here, take $G=C_{12}=\langle x\rangle$ be cyclic of order $12$, and take $a_1=x^2$, $a_2=x^3$. Then $\{a_1,a_2\}$, and no proper subset of $\{a_1,a_2\}$ generates. The order of $a_1$ is $6$, the order of $a_2$ is $4$, but $x^6$ can be expressed as either $(x^2)^3(x^3)^0$, or as $(x^2)^0(x^3)^2$, so the representation is not unique.
Interpretation 2. $G=\langle a_1,\ldots,a_n\rangle$, and if $X$ is any set with $\langle X\rangle = G$, then $|X|\geq n$; that is, $\{a_1,\ldots,a_n\}$ generates, and no set of cardinality strictly smaller than $n$ generates.
This time take $G=C_2\times C_4$, with $x$ a generator of order $2$ and $y$ a generator of order $4$ with $\langle x\rangle\cap\langle y\rangle =\{e\}$. Now let $a_1=xy$ and $a_2=y$. This is a generating set of smallest possible order; $|a_1|=|a_2|=4$. But $y^2$ can be expressed both as $(a_1)^2(a_2)^0$ and as $(a_1)^0(a_2)^2$. So the representation is not unique either. 
(The second example also works for the first interpretation, but the failure in the first interpretation is much more radical, since you can have really big “minimal generating sets” even for groups that can be generated by very few elements.)
A: 2.) Suppose there existed a homomorphism from $D_4$ to $\mathbb{C}_8$. Then $b\in D_4$, which has a order of $2$ and is a generator of $D_4$ must be mapped to an element of order $2$. In $\mathbb{C}_8$, the only element which has order $2$ is the element $x^4 (\text{if } \mathbb{C}_8 = \langle x\rangle)$. So this homomorphism maps $b \rightarrow x^4$. Similarly, $a$ must have an order a divisor of $4$, i.e. $2$ or $4$. If it was mapped to an element of order 4, it would lead to a contradiction, which you can try and see. Map $a$ to $x^4$ as well. You will notice that this is homomorphism such that is not trivial. 
A: For the second question, any such homomorphism would have a kernel, a normal subgroup of $D_4$. There are three normal subgroups: $\langle e\rangle,\langle a^2\rangle,\langle a\rangle$.
The quotient by $\langle e\rangle$ is all of $D_4$, which does not fit inside $C_8$ for obvious reasons.
The quotient by $\langle a^2\rangle$ is the Klein 4-group, which does not fit inside $C_8$ because it has three elements of order $2$.
The quotient by $\langle a\rangle$ is $C_2$, which fits inside $C_8$ as $\langle x^4\rangle$ (where $x$ is a generator of $C_8$).
So map $a\mapsto e$ and $b\mapsto x^4$.
