# Group on Five Elements

• Jan 21st 2010, 04:02 PM
Haven
Group on Five Elements
I am having troubles proving this:
Prove that any group on five elements is cyclic.
• Jan 21st 2010, 04:07 PM
Drexel28
Quote:

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

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

Your example is just a corollary.
• Jan 21st 2010, 07:18 PM
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 $x \in G$ such that $x \neq e$.
so $|x| \in \{2,3,4,5\}$
if $|x| = 5$, then we are done, since $G = $.

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

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
• Jan 21st 2010, 07:20 PM
Drexel28
Quote:

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 $x \in G$ such that $x \neq e$.
so $|x| \in \{2,3,4,5\}$
if $|x| = 5$, then we are done, since $G = $.

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

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!
• Mar 20th 2010, 05:05 AM
elements
Quote:

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
• Mar 20th 2010, 05:40 AM
proscientia
Quote:

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 $g\in G$ and $g\ne e,$ then $\langle g\rangle=G$ – which is to say that $g$ has order 5.