Question

Prove that a group of order 33 contains an element of order 3.

Short answer

It is proved that, a group of order 33 contains an element of order 3.

Step-by-step solution

Determine that a group of order 33 contains an element of order 3

Consider that G doesn’t have any elements of order 3.

Then, by using Lagranges theorem, each non-identity element of G have order 11 or 33.

If there is an element $a\in G$and the order of a is 33, then G is a cyclic group which is generated by a.

Consequently, the element $a^{11}\in G$has order 3.

$\begin{array}{rcl}{\left(a^{11}\right)}^3& =& a^{11+11+11}\\ & =& a^{33}\\ & =& e\\ & & \end{array}$

Further simplification

Now, let’s suppose that G is not cyclic.

Then, every non-identity element of G has order 11.

Taking element such as, $x_1,x_2,x_3\in G$then the cyclic subgroups is .

By the property and $x_j\in G/H_{J-1}j=2,3$, now, claim that for $i,j=1,2,3$with $i\ne j$.

$H_i\cap H_j=\left\{e\right\}$

Here,e is the identity element of G.

Consider that $g\in H_i\cap H_j$with $g\ne e$for$i\ne j\in \left\{1,2,.....,n\right\}$ .

Then,$g={x_i}^k=x_j^l$ .

For $0⩽k,l⩽p-1$implies that, $x_i={x_j}^{l+n-k+1}\in H_j$.

This is not possible from the choice of $x_i,i=1,2,3$.

Hence, claim is proved.

Further simplification

The cardinality of $H_1\cup H_2\cup H_3$is 31.

Thus, there is an element $g\in G/H_1\cup H_2\cup H_3$and from the hypothesis, the order is also 11.

As order of G is 33, there exists$2⩽j⩽10$ so that $g^j\in H_i$for .$i=1,2,3$

Since in this case it yields then, $g^j=x_i^l$.

For$1⩽l⩽10$ implies that the order of $g^j$is 11 as each element of$H_i$ has the order 11.

This cannot happen since the order of g is 11j.

Therefore, the order of g has to be 3.

Hence proved.