0

Trying to understand the group order material. While practising, I came across with the following question.

Consider $g\in G$ so $o(g)=15$. calculate $o(g^7),o(g^{-1})$.

I'm not sure how to approach this question. I understand from $o(g)=15$ that $g^{15}=e$. So in order to calculate $o(g^5)$ we will have to do some arithmetic on $g^{15}$. we need to find $k\in\mathbb{N}$ so $(g^7)^k = e$. We are getting $g^{7k}=e$ and now I'm not sure what do next. Also probably the bigger problem is to understand how to calculate $o(g^{-1})$.

vesii
  • 1,979

2 Answers2

2

The assertion $\operatorname{ord}(g)=12$ means two things:

  1. $g^{12}=e$;
  2. if $k\in\{1,2,\ldots,11\}$, then $g^k\neq e$.

A standard fact about the order of an element is this: $g^k=e\implies\operatorname{ord}(g)\mid k$.

From the fact that $g^{12}=e$, you deduce that $(g^5)^{12}=e$, since$$(g^5)^{12}=g^{5\times12}=(g^{12})^5=e^5=e.$$Therefore, $\operatorname{ord}(g^5)$ is at most $12$.

Now, take $k\in\{1,2,\ldots,11\}$ and assume that $(g^5)^k=e$. This means that $g^{5k}=e$. But then, since $\operatorname{ord}(g)=12$, $12\mid5k$. Since $\gcd(12,5)=1$, it follows that $12\mid k$, which is impossible, since $k\in\{1,2,\ldots,11\}$. This proves that $\operatorname{ord}(g^5)=12$.

Now, prove that $\operatorname{ord}(g^{-1})$ is also equal to $12$.

José Carlos Santos
  • 427,504
0

If you are concerned with the order of $g^{-1}$ ("probably the bigger problem"), then everything is fine. It is easy to see that $o(g)=o(g^{-1})$:

An element of a group has the same order as its inverse

For the first question, by Lagrange, the order of the subgroup generated by an element divides the order of the group.

Dietrich Burde
  • 130,978