Question

Suppose G has order 4, but contains no element of order 4.

Prove that no element of G has order 3.

Short answer

It is proved that, no element of G has order 3.

Step-by-step solution

Referring to the Hint

Consider G=$\mathbf{\{}\mathbf{1}\mathbf{,}\mathbf{g}\mathbf{,}\mathbf{\text{\hspace{0.17em}}}{\mathbf{g}}^{\mathbf{2}}\mathbf{,}\mathbf{d}\mathbf{\text{\hspace{0.17em}}}\mathbf{\}}$ where d is order of 3 elements.

Given that G has order 4, but contains no element of order 4

Proving that no element of G has order 3

As we consider G=$\{e,g,\text{\hspace{0.17em}}{g}^2,d\text{\hspace{0.17em}}\}$ where d is an order of 3 elements.

Since we know that G does not have order 4 elements, gd cannot be an element in G. Now take each case and verify our assumption as follows:

1. Let gd =e.

Then, $d=g^{-1}\Rightarrow g^{3}d=g^2$.

Since $g^2$is a group element, it contradicts our assumption.

2.

Let gd= g.

Then, d = e.

Since e is a group element, it contradicts our assumption.

3.

Let $gd=g^2$.

Then, d=g.

Since g is a group element, it contradicts our assumption.

4.

Let gd=d.

Then g=e.

Since e is a group element, it contradicts our assumption.

Therefore, there is a contradiction in all the above cases.

Hence, it is proved that G does not have an element of order 3.