Find pair of product of four groups that has the same order, but not isomorphic.

Question

Find a set of $16$-element product of four groups that have same number of elements of each order, but are not isomorphic.

Choices to form two equal order products of four groups: $C_1, C_2, C_4, C_8, C_{16}, Q_8.$

---

Need find products of four groups, and to form such two products having equal order, but not isomorphic.

Le, the first product of four groups be denoted by $G= a\times b\times c\times d$, and the second by $H= e\times f\times g\times h.$

The choice need to show equal order, while failing at Isomorphism.

So, $G_1= C_{16}\times C_1\times C_1\times C_1$, and $H= C_8\times C_8\times C_1\times C_1$ will not help; as then will have equal orders as well as Isomorphism too.

A possible choice is : $G_1= C_{8}\times C_1\times C_1\times C_1$, and $H= Q_8\times C_1\times C_1\times C_1$

Another choice: $G_1= C_{16}\times C_1\times C_1\times C_1$, and $H= Q_8\times C_8\times C_1\times C_1$

But, the answer key states: On writing group product, in decreasing group order, have only one choice for determining $G,H$ of order $16$.

There should be many choices possible, as not specified to use a group only one time in product. Seems it is implicit to use a group only one time in a product.

[cancel]{Edit:. One possible answer is $$G: Q_8\times C_2\times C_1\times C_1,$$ $$ H: C_{16}\times C_4\times C_2\times C_1,$$ both of order $|G|=|H|=16.$ But, there should be many choices possible, unless each group can be used once in a product.}[/cancel]

Final Edit: Seems with restriction on both product groups having the same number of elements order-wise, it is impossible to state even the factors of each product. Getting two products (or, even one ), of length $16$ from given set of values is impossible.

Answer 1

So,$ G_1=C_{16}×C_1×C_1×C_1$ and $H=C_8×C_8×C_1×C_1 $will not help; as then will have equal orders as well as Isomorphism too.

Both groups neither have the equal order nor they are isomorphic. $G_1$ has an element of order $16$ but the largest order of an element of $H$ is $8$ . In fact $G_1\cong C_{16}$ and $H\cong C_8×C_8$

( one is cyclic but other is not!)

Key results:

1. $G×G'$ is abelian iff $G, G'$ both are abelian.

2. $C_m ×C_n\cong C_{mn}$ iff $(m, n) =1$

3. $C_8\ncong Q_8$

---

Now you can produce required example.

$G=C_2×C_4×C_8×C_{16}$

$H=C_2×C_4×Q_8×C_{16}$

They are not isomorphic any more as $H$ is non abelian but $G$ is abelian.

Answer 2

How about $C_2 \times C_4 \times C_8 \times C_{16}$ and $C_2 \times C_4 \times Q_8 \times C_{16}$? The first group is abelian, the second is not.