It can be shown by constructing all possible multiplication

It can be shown by constructing all possible multiplication tables that a group of order 4 must either be isomorphic to Z₄ or Z₂ Z₂. Use this to show that all proper subgroups of D₄ are Abelian.

Solution

It is known that, D4 has 8 elements: 1, r, r² , r³ , d₁, d₂, b₁, b₂,

Thus, |D4| = 8.

Thus, by a theorem of Lagrange, we know that any proper subgroup has an order which factors 8. Thus if it were possible to accomplish this decomposition, one group must have order 4 and the other order 2. Any group of order 4 is abelian and \"the\" group of order 2 is abelian. The product of two such groups would also be abelian. Therefore D4 would be isomorphic to an abelian group and would therefore be abelian.

However, we know that D4 is not abelian. Thus we have a contradiction and consequently D4 cannot be expressed as the internal direct product of its proper subgroups.

further, In the case of D4 the only non-cyclic groups besides D4 itself can only be of order 4.

So,we have subgroups of G that consist of the identity, and three involutions (elements of order 2).

The normal subgroups of D4 are D4 and {1}, and all the elements of D4 are either of order 2 or 4 and we know that

groups of order 4 are isomorphic to either Z4or Z2 Z2. So, all the proper subgroups of D4 are abelian.