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I have a question that says this:
Classify $\mathbb{Z}_5 \times \mathbb{Z}_4 \times \mathbb{Z}_8 / \langle(1,1,1)\rangle$ according to the fundamental theorem of finitely generated abelian groups.
I would like to see how it is correctly answered. This is not homework; I'd just like to see a proof.
To make this a more or less computational manner, consider $$ \mathbb Z_5\times\mathbb Z_4\times\mathbb Z_8\big/\left\langle(1,1,1)\right\rangle \cong \mathbb Z^3\big/U, $$ where $$ U= \left\langle \pmatrix{5\\0\\0}, \pmatrix{0\\4\\0}, \pmatrix{0\\0\\8}, \pmatrix{1\\1\\1} \right\rangle. $$ Now our goal is to find bases of $U$ and $\mathbb Z^3$ (both as $\mathbb Z$-modules) that reveal the structure of the quotient. Use the fact that $\gcd(5,8)=1$ and $8=2\cdot 4$ to obtain $$ U= \left\langle \pmatrix{1\\0\\0}, \pmatrix{0\\4\\0}, \pmatrix{0\\1\\1} \right\rangle. $$ Where those $3$ elements of $\mathbb Z^3$ are now $\mathbb Z$-linearly independent. Now we pick a similar basis of $\mathbb Z^3$, namely $$ \mathbb Z^3 = \left\langle \pmatrix{1\\0\\0}, \pmatrix{0\\1\\0}, \pmatrix{0\\1\\1} \right\rangle. $$ Thus, the quotient kills two of the basis elements and leaves us with $$ \mathbb Z^3\big/U \cong \langle(0,1,0)\rangle\big/\langle(0,4,0)\rangle \cong \mathbb Z_4. $$
In the end, this is just linear algebra over $\mathbb Z$, which is a little bit more tricky since we can't divide, but you get the gist of it. You might be interested in Smith normal forms, which basically tells you how to choose bases for this to work.
I'm assuming that $\langle(1,1,1)\rangle$ means the subgroup generated by $(1,1,1)$, or in other words $\{(k\bmod 5,k\bmod 4,k\bmod 8)\mid k\in\mathbb Z\}$.
In that case we can see that each of the cosets that make up the quotient must contain an element of the form $(0,x,0)$. Namely, assume that $(a,b,c)$ is some element of the cosets; then by the Chinese Remainder Theorem we can find $k$ such that $a\equiv k\bmod 5$ and $c\equiv k\bmod 8$. Subtracting $(a,b,c)-(k,k,k)$ gives us an alement o the form $(0,x,0)$.
On the other hand, $(0,x,0)$ is can only be zero in the quotient when $x\equiv 0\bmod 4$ (because if $(0,x,0)\equiv(k,k,k)$ then $k\equiv 0\bmod 8$ and therefore $k\equiv 0\bmod 4$.
So the quotient group is $\mathbb Z_4$.
What this has to do with the structure theorem I don't know, though. Perhaps you're supposed to start by rewriting it to $\mathbb Z_4\times\mathbb Z_{40}/\langle(1,1)\rangle$?
For convenience, let $\mathcal G = \mathbb{Z}_5 \times \mathbb{Z}_4 \times \mathbb{Z}_8 / \langle(1,1,1)\rangle$
Since $\text{ord}\langle(1,1,1)\rangle = 40$, then $\text{ord} (\mathcal G) = \dfrac{5 \times 4 \times 8}{40} = 4$
Using this we compute
a: 5 4 8
s: 1 1 1
g: 1 1 1
m: 5 4 8
p: 5 1 8
t: 1 4 1
r: 1 4 1
α: 1 0 0
ω: 1 1 1
β: 1 0 0
And conclude that every $(x,y,z) \in \mathcal G$ can be expressed uniquely as
$$(x,y,z) = \kappa(1,1,1) + \lambda(0, 1, 0)$$
where $0 \le \kappa \lt 40$ and $0 \le \lambda \lt 4$
Consequently, $\mathcal G \cong \mathbb Z_4$
The Smith Normal Form of the matrix $\begin{pmatrix} 5&0&0\\ 0&4&0\\0&0&8\\1&1&1\\\end{pmatrix}$ (which gives a presentation for the group) has diagonal entries 1, 1, 4; thus the group is isomorphic to $Z_4$.
