This is in continuation of earlier post here. Though still not found two such $16$-length product groups, still want to learn the process of finding correct pair of such product groups.
Goal is to show that two $16$- length groups formed by four groups' products, are not isomorphic; though they have same number of elements order-wise.
Let, $G= C_{8}\times C_2\times C_1\times C_1$, and $H= Q_8\times C_2\times C_1\times C_1$
First, need to know if both $G,H$ have $16$- length products.
Next, need show two products $G,H$ have the same number of elements order-wise.
Last part is to show these two groups are not isomorphic, i.e. $G\ncong H.$
Request inputs for first part, as assume that $C_1$ has only identity element that does not add to length of group product.
For second part, my attempt:
So, Cartesian product need be formed
$G$ has components $C_8, C_2, C_1.$
Number of elements of order $8$ in $C_8=2,$ given by $r, r^7,$
Number of elements of order $4$ in $C_8=2,$ given by $r^2, r^6,$
Number of elements of order $2$ in $C_8=1,$ given by $r^3,$
Number of elements of order $1$ in $C_8=1,$ given by $r^0=e.$
Second consider $C_4$:
Number of elements of order $4$ in $C_4=2$ given by $r_4, r_4^3,$
Number of elements of order $2$ in $C_4=1,$ given by $r_4^2,$
Number of elements of order $1$ in $C_8=1,$ given by $r_4^0=e.$
Now, form Cartesian product of two groups : $$C_8=\{e_{C_8}, r_{C_8}, r_{C_8}^2, r_{C_8}^3, r_{C_8}^4, r_{C_8}^5, r_{C_8}^6, r_{C_8}^7\},$$ $$C_4:\{e_{C_4}, r_{C_4}, r_{C_4}^2, r_{C_4}^3, $$ $$CP_{C_8\times C_4}= \{(r_{C_8}^0, r_{C_4}^0), (r_{C_8}^0, r_{C_4}^1), ..., (r_{C_8}^7, r_{C_4}^3) \}$$
But $e$ can pair only with other $e$.
So, get :
$$CP_{C_8\times C_4}= \{(r_{C_8}^0=e_{C_8}, r_{C_4}^0=e_{C_4}), (r_{C_8}, r_{C_4}), (r_{C_8}^1, r_{C_4}^2), (r_{C_8}^1, r_{C_4}^3), $$ $$(r_{C_8}^2, r_{C_4}), (r_{C_8}^2, r_{C_4}^2), (r_{C_8}^2, r_{C_4}^3),$$ $$(r_{C_8}^3, r_{C_4}), (r_{C_8}^3, r_{C_4}^2), (r_{C_8}^3, r_{C_4}^3), ... \}$$ Or, a total of ($3\times 7+1= 22$) elements in the Cartesian product.
But, need know order of these elements pair-wise.
Seems a long exercise, is there a shortcut?
For last part, think if above is correct, then can use the property that $C_8\ncong Q_8$.
The direct product of two abelian groups is abelian. Thus
$$G= C_{8}\times C_2\times C_1\times C_1$$
is abelian. However, since $Q_8$ is nonabelian, $Q_8\times A$ is nonabelian for any group $A$. Thus
$$H= Q_8\times C_2\times C_1\times C_1$$
is nonabelian.
For any groups $K$ and $L$ with $K\cong L$, we have that $K$ is abelian if and only if $L$ is abelian.
Thus the groups $G$ and $H$ cannot be isomorphic.