# Math Help - Order of a group

1. ## Order of a group

I got the following from Wikipedia:

The order of a group is its cardinality, i.e., the number of its elements;
the order, sometimes period, of an element a of a group is the smallest positive integer m such that am = e (where e denotes the identity element of the group, and am denotes the product of m copies of a). If no such m exists, we say that a has infinite order. All elements of finite groups have finite order.

Question: If I have $a^m$ where $m=4$, is it correct to say that $|G|=4$. If so, then $G$contains $4$ elements. Must all the elements be distinct? Must an identity be in the group?

2. ## re: Order of a group

"Question: If I have where , is it correct to say that . If so, then contains elements. Must all the elements be distinct? Must an identity be in the group"

If you mean that G is a cyclic group (in other words, the entire group is $G = \{a^0, a^1, a^2, a^3, a^4 \}$ ) such that $a^4$ is as high as you can get without repeating, then actually $|G| = 5$ because $G$ has 5 elements (count them).

Yes, by definition, a group must have an identity, which in this case is $a^0$, and all elements must be distinct. For example, although $a^0$ and $a^5$ look different, they're actually the same element because in a cyclic group, the elements start repeating.''

3. Since $m=4$ and $|G|=5$, is it of the order of $4$ or $5$?

Drawing from what is said in Wikipedia, when $m$ is the smallest positive integer such that $a^m=e$, then $m=0$ must be the number, then what is the order? hmm?

4. In my example, since $|G| = 5$, the order of the group is 5. But what exactly do you mean when you say, "I have $a^m$ where $m = 4$"? We can raise $a$ to any power we want and the result will always be in the group $G$. However, if this is a finite group, we'll get a lot of repetition.

Let's say you start out with $a$, raise it to some power $n$, and get the same thing as the identity, $a^0$. If $n$ is the smallest positive integer power such that $a^0 = a^n$, then $n$ is the order of the group. In my example, since 5 is the smallest power that gives me back the identity, that's the order of the group.

Thus, if I have a group of order $n$, since I don't want duplicates in my listing, I'll only raise $a$ to the powers $0, 1, ..., n-1$. This will give me every unique element in the group.

5. Your explanation makes sense. Thanks.

There is an example at wikipedia Order (group theory) - Wikipedia, the free encyclopedia

It shows a symmetric group table. In the example, each element $s,t$, and $w$ square to $e$. Then it says that these group elements have order 2.

I am a little confused by the terms "order of a group" and "order of group elements." Are they the same?

6. Nope, they're not the same--it's confusing at first, I know. The order of a group'' G is the number of elements it contains. Consider $G = \{e, a, a^2, a^3 \}.$ The order of this group is 4 since it has 4 elements.

The order of an element'' $a$ means: to what smallest power do I have to raise this element to get back the identity? Consider the element $a^2$ in the group I described above. I know
$(a^2)^2 = a^4 = e$. Thus, the element $a^2$ has order 2. So you see that the order of a group and the order of an element are not the same, but in fact, the order of an element always divides the order of a group (notice that 2, the order of $a^2$, divides 4, the order of the group.)

In your example from Wikipedia, it says if you square those elements, you get the identity. By definition, then, these elements have order 2. Make sense?

7. Since $s,t$, and $w$ square into $e$, which means that $s^2=s*(s^{-1})=e$, similarly $t$ and $w$. If $s=s^{-1}=e$, then it's trivial. Now, if $s \not =s^{-1}$, how do you explain $s^2=e$? What is the simplest example?

8. So we're assuming that $s^2 = e,$ which implies $s = s^{-1}$ (you can see this by multiplying each side of the first equation by $s^{-1}$). Thus, you cannot have both $s^2 = e$ and $s \not= s^{-1}$--it's a contradiction.

If an element $s$ has order 2, it is its own inverse, i.e. $s = s^{-1}$. Certainly, if $s = e$, then $s^2 = e$. But $s$ doesn't have to be the identity for this to happen.

Let $G = \{e, a, a^2, a^3, a^4, a^5 \}$. Then consider what happens when we square $a^3$. We'll have $(a^3)^2 = a^6 = e$. So we know $a^3 = (a^{3})^{-1}$ in this group, following similar reasoning to the first paragraph.

So in summary, if $s$ squares to $e$, you definitely know that $s = s^{-1}.$ But $s$ doesn't necessarily have to be the identity.

9. In terms of a real number, if $\forall a \in \mathbb{R}, a^0=1$.

What will $a$ be where $a\not \in \{0,1\}$, $a=a^{-1}$ and $a^{2n}=1$, $n \in \mathbb{N}$? I can't think of any. Can it be an imaginary number?

$-i^2 = 1$

10. If "Multiplication" is the group operation, then we're treating 1 as the identity for this group. Keep in mind, this means 0 is not the identity for this group. (0 is the identity for a different group, where "addition" is the operation)

What about $a = -1$? We know $(-1)^{2n} = ((-1)^2)^n= 1$, regardless of what $n$ is. Since $(-1)^2 = 1$, from previous discussion we know $-1 = (-1)^{-1}$. So $a = -1$ satisfies the criteria, doesn't it?

In fact $a$ must be a real number since when you square a complex number, you double its angle (measured counter-clockwise about the origin from the real axis). So the only possibility is for $a$ to have angle 180 (so that it's a negative number) or angle 0 (so that it's a positive number).

11. Sir, I want to thank you for your time for answering all my questions. You gave very clear instructions at every step.