Vtyjkyuk

What group is isomorphic to $(mathbbZ_4timesmathbbZ_4timesmathbbZ_8)/langle(1,2,4)rangle$?

I can only see that $mathbbZ_4timesmathbbZ_4timesmathbbZ_8$ has $128$ elements and $langle(1,2,4)rangle$ has $4$ elements, so this factor group will have $128/4=32$ elements, and will therefore be isomorphic to exactly one of the following groups:
$$
mathbbZ_32,
quad
mathbbZ_16timesmathbbZ_2,
quad
mathbbZ_8timesmathbbZ_4,
quad
mathbbZ_8timesmathbbZ_2timesmathbbZ_2,
quad
mathbbZ_4times mathbbZ_4timesmathbbZ_2
$$

However, determining which one it is actually isomorphic to seems like a messy work, I can only do it by writing all the cosets and see which abelian group it is isomorphic to. Is there any faster and reliable method?

As explained in detail here you need to find the Smith normal form of the matrix $$pmatrix4&0&0\0&4&0\0&0&8\1&2&4$$
First, use the $1$ in the bottom-left to get rid of the last row and first column:
$$pmatrix4&0&0\0&4&0\0&0&8\1&2&4topmatrix4&-8&-16\0&4&0\0&0&8\1&0&0topmatrixcolorred0&-8&-16\colorred0&4&0\colorred0&0&8\colorred1&colorred0&colorred0$$
The isolated $1$ corresponds to a factor $Bbb Z/1Bbb Z$ which is the trivial group, so we forget about it. We proceed with the remaining $3times 2$ matrix
$$pmatrix-8&-16\4&0\0&8to pmatrixcolorred0&-16\colorred4&colorred0\colorred0&8$$
where a factor $Bbb Z/4Bbb Z$ splits out.

Finally $$pmatrix-16\8topmatrix0\8$$ gives you the last factor, $Bbb Z/8Bbb Z$. Hence the quotient is isomorphic to: $$Bbb Z/4Bbb Ztimes Bbb Z/8Bbb Z$$