# Order of an element and $\mathbb{Z_n}$

Consider the group $$\mathbb{Z_8}^*$$. This group contains only the following elements, 1,3,5,7. It was suggested to me that this group is not cyclic because there does not exist an element in the set with order 4.

I do not see how I can prove that for an element g in $$\mathbb{Z_n}^*$$ the integer n in $$g^n$$ gives me the number of elements in the set that is generated by g.

My thoughts on this are as follows: I believe that it may be the case that the order of an element in a group g tells me the number of elements in the generator of g, because G is closed and the order is the smallest positive integer n that gives the identity.

• Are you asking only if the order of a cyclic group equals the order of its generator, or something further? That surely has been asked in the past so it will end up being a duplicate question (always search first before asking). – Bill Dubuque Feb 2 '19 at 18:42
• Well, you know $|Z_8^*|=4$ because you can count the elements. 1-> one, 3-> two, 5-> three, 7-> four. There are four of them. So if $Z_8^*$ is cyclic then there is a $g$ so that $\{g, g^2, g^3, g^4, g^5,.........\} = \{1,3,5,7\}$. Since $\{1,3,5,7\}$ has four different elements then $\{g,g^2,g^3,g^4,g^5....\}$ must have four differen elements. So $\{g,g^2,g^3,g^4,g^5....\} = \{g,g^2,g^3,1\}$ and $g^4 = 1$ and $g^2 \ne g$ and $g^2 \ne 1$. and $g^3 \ne 1$. and $g^3 \ne g^2$. And it's just a matter of testing all the elements. There is no such element. – fleablood Feb 2 '19 at 19:22

By Lagrange's Theorem we know that for any group $$G$$ with $$\left| G \right| = n$$ we have $$g^n = 1$$ for any $$g \in G$$. If the group $$G$$ is cyclic then there must be an element $$g$$ such that $$g$$ has order $$n$$ (by definition of cyclicity). So if you check that no element has order $$4$$ for $$\mathbb{Z}_8^*$$ then the group is not cyclic. Does this help?

I think what you are asking is if $$g\in G$$ and $$g^k = 1$$ and that $$k$$ is the smallest positive integer, $$k$$ so that $$g^k =e$$ (i.e. the definition $$|g| = k$$) then $$|| = k$$.

Pf: $$ = \{g, g^2, g^3,......., g^l,.....\}$$

If $$m > k$$ then $$g^m = g^{k + (m-l)} = g^k*g^{m-1} = e*g^{m-1} = g^{m-l}$$ so $$g^m$$ is not distinct and $$$$ will have at most $$k$$ elements.

And $$ = \{g, g^2, g^3, ....., g^{k-1}, g^k=e\}$$.

Now suppose those elements are not distinct, that is, suppose $$g^i = g^j$$ but $$0 < i < j < k$$.

The $$g^i = g^j$$

$$g^i*(g^{-1})^i = g^j*(g^{-1})^i$$

$$e = g^{j-i}$$.

But $$0 < j-i < k$$ and that contradicts that $$k$$ was the smallest such positive integer where $$g^k = e$$.

So $$ = \{g, g^2, .... ,g^{k-1}, 1\}$$ and $$|| = k$$.

...

That was probably already proven in your class and is buried somewhere in your notes. You definitely used that when you proved Lagrange Theorem.

Just do it.

If $$\mathbb Z_8^*$$ is cyclic there is a $$g\in \mathbb Z_8^*$$ so that $$\{g,g^2,g^3,g^4\} = \{1,3,5,7\}$$

$$\{1,1^2, 1^3, 1^4\} = \{1,1,1,1\} = \{1\} \ne \{1,3,5,7\}$$

$$\{3, 3^2, 3^3, 3^4\} = \{3,1,3,1\} = \{1,3\} \ne \{1,3,5,7\}$$

$$\{5, 5^3, 5^3, 5^4\} = \{5,1,5,1\} = \{1,5\} \ne \{1,3,5,7\}$$

$$\{7,7^2, 7^3, 7^4\} = \{7,1,7,1\} = \{1,7\}\ne \{1,3,5,7\}$$

So there is no such $$g$$ so it isn't cyclic.

I do not see how I can prove that for an element g in Zn∗ the integer n in gn gives me the number of elements in the set that is generated by g.

Assuming $$n$$ is the same as the index i $$\mathbb Z_n$$ then $$g^n =1$$. This is a direct consequence of Lagrange theorem: If $$G$$ is a finite group of order $$n$$, then $$|g|=$$ divides $$|G| = n$$. And so $$g^n = (g^k)^{\frac nk} = 1^{\frac nk} = 1$$.

My thoughts on this are as follows: I believe that it may be the case that the order of an element in a group g tells me the number of elements in the generator of g, because G is closed and the order is the smallest positive integer n that gives the identity.

Well that's obvious.

If $$|g| = k$$ then BY DEFINITION $$g^k = 1$$ and $$g^{i; 0 so $$g, g^2,g^3, g^4, ......, g^k =1$$ will be the element in $$$$ and if you count them there are $$k$$ of them.

Okay, we have to prove then $$g^i \ne g^j$$ if $$0 \le i < j < k$$ but that's clear: If $$g^i = g^j$$ then $$g^i*(g^{-1})^i = g^j*(g^{-1})^i$$ so $$1 = g^{j-i}$$ but $$0 < j-i < k$$. But that contradicts $$|g| = k$$.