# Thread: Group on Five Elements

1. ## Group on Five Elements

I am having troubles proving this:
Prove that any group on five elements is cyclic.

2. Originally Posted by Haven
I am having troubles proving this:
Prove that any group on five elements is cyclic.
Do you know Lagrange's theorem?

Lemma: If a set has no proper non-trivial subgroups, then it's cyclic.

Proof: Let $\displaystyle G$ meet the criterion. Let $\displaystyle g\in G-\{e\}$ be arbitrary. We must then have that $\displaystyle \left\langle g\right\rangle$ is a subgroup of $\displaystyle G$ that is non-trivial. Thus, by assumption we must have that $\displaystyle \left\langle g\right\rangle$ is improper, thus equal to $\displaystyle G$. The conclusion follows. $\displaystyle \blacksquare$

Now, using this lemma and Lagrange's theorem we see that any subgroup of $\displaystyle G$ where $\displaystyle \left|G\right|=p$ for some prime $\displaystyle p$ must be a divisor of $\displaystyle p$. Thus, any subgroup must be of order $\displaystyle 1,p$. Therefore, the lemma implies that $\displaystyle G$ is cyclic for any prime-ordered group.

Your example is just a corollary.

3. I'm supposed to prove the statement by arguing that all the non-identity elements of G have order 5.

So far I have
Let $\displaystyle x \in G$ such that $\displaystyle x \neq e$.
so $\displaystyle |x| \in \{2,3,4,5\}$
if $\displaystyle |x| = 5$, then we are done, since $\displaystyle G = <x>$.

Case 1
$\displaystyle |x| = 4$
Then $\displaystyle <x> = \{e,x,x^2,x^3\}$ and $\displaystyle G = \{e,x,x,x^3,y\}$ where $\displaystyle y \notin <x>$.
If $\displaystyle xy = x,y \rightarrow x,y=e$
If $\displaystyle xy = x^2,x^3 \rightarrow y = x,x^2$ respectively
If $\displaystyle xy = e \rightarrow y=x^3$

Since they are all contradictory, x cannot have order 4. So no element in G can have order 4.

But I can't figure out the other cases

4. Originally Posted by Haven
I'm supposed to prove the statement by arguing that all the non-identity elements of G have order 5.

So far I have
Let $\displaystyle x \in G$ such that $\displaystyle x \neq e$.
so $\displaystyle |x| \in \{2,3,4,5\}$
if $\displaystyle |x| = 5$, then we are done, since $\displaystyle G = <x>$.

Case 1
$\displaystyle |x| = 4$
Then $\displaystyle <x> = \{e,x,x^2,x^3\}$ and $\displaystyle G = \{e,x,x,x^3,y\}$ where $\displaystyle y \notin <x>$.
If $\displaystyle xy = x,y \rightarrow x,y=e$
If $\displaystyle xy = x^2,x^3 \rightarrow y = x,x^2$ respectively
If $\displaystyle xy = e \rightarrow y=x^3$

Since they are all contradictory, x cannot have order 4. So no element in G can have order 4.

But I can't figure out the other cases
Jeez. You basically have to construct the group table....good luck!

5. Originally Posted by Haven
I am having troubles proving this:
Prove that any group on five elements is cyclic.
Fascinating. I was looking up the Taoist 5 Elements and I found your thread about mathematical 5 elements.

Not being a mathematician, I don't get it. Would you mind giving me the basic premise of the mathematical 5 elements concept?

In Taoism, any five cyclical system is represented by wood, fire, earth, metal, and water. How each supports each other is obvious - back down to water, the fifth element supporting wood, the first element all over again. This characterization is supposed to be a perfect analogy for anything set of five that are mutually supportive. I found a fuller description of the 5 elements

6. Originally Posted by Haven
I'm supposed to prove the statement by arguing that all the non-identity elements of G have order 5.
But that was exactly was Drexel28 did! Drexel28 proved that if $\displaystyle g\in G$ and $\displaystyle g\ne e,$ then $\displaystyle \langle g\rangle=G$ – which is to say that $\displaystyle g$ has order 5.