# Show only one group exists for Groups of prime order

• May 29th 2010, 06:30 PM
GeoC
Show only one group exists for Groups of prime order
I'm trying to dust off the cobb webs by studying some basic Group Theory. Can someone provide a proof that a Group of order 5, or any prime for that matter, must be a cyclic Abelian group, and that there can be only one such group...

I can easily find the multiplication table for the Group, but I don't see how to prove the statement that it is the only such group.

Any insights would be much appreciated..

Thanks
• May 29th 2010, 07:17 PM
roninpro
Hi there.

You might try thinking about Lagrange's Theorem, which states that if $G$ is a finite group, the order of any subgroup $H$ must divide the order of $G$. Can you see how to use this to show that every group of prime order is cyclic?
• May 29th 2010, 10:13 PM
Drexel28
Quote:

Originally Posted by roninpro
Hi there.

You might try thinking about Lagrange's Theorem, which states that if $G$ is a finite group, the order of any subgroup $H$ must divide the order of $G$. Can you see how to use this to show that every group of prime order is cyclic?

And of course then use this and the important fact that every cyclic group is isomorphic to $\mathbb{Z}_n$ or $\mathbb{Z}$ to finish.
• May 30th 2010, 05:03 AM
GeoC
Quote:

Originally Posted by roninpro
Hi there.

You might try thinking about Lagrange's Theorem, which states that if $G$ is a finite group, the order of any subgroup $H$ must divide the order of $G$. Can you see how to use this to show that every group of prime order is cyclic?

I see that a group of prime order cannot have any subgroups (other than E), but how does that lead to the non-existance of groups of a given prime order, other than the cyclic, Abelian group? For example, I can create more than one multiplication table for $G_5$ which obeys the rule that no element appear more than once in any row or column. However, I can show the table for the non-Abelian $G_5$ violates the associative property, and thus is not a proper group. But such a brute force approach is not practical for primes of higher order.

I guess the cobb webs have pretty high tensile strength!
• May 30th 2010, 05:34 AM
GeoC
Am I heading in the right direction:

If the order $n$ of any element $g$ of the group $G_p$, where $p$ is the order of the group $G$, is such that $g^n=E$, and by Legrange's theorem $n$ must be divisible into $p$, then for $p$ prime, it must be true that $n$ is either 1 or $p$.

Since every element $g_k$ of the Group $G_p$ must have an order $n$, and since $n$ is restricted to either 1 or $p$ when $p$ is prime, all elements (except E) must have the same order -- i.e. $p$.

Now, need to show that only one such Group $G_p$ can satisfy this condition.

??
• May 30th 2010, 10:20 AM
roninpro
You're on the right track. Every element $g$ in a group of order $p$ has order 1 or $p$. What can be said about the subgroup generated by $g$, if $g$ is not the identity?
• May 30th 2010, 12:45 PM
GeoC
If g is not the identity, and it has order p, then $g^1, g^2, g^3, ..... g^p$ should generate all the elements of the group. Thus, I believe I can say three things about the subgroup:

1) The subgroup generated by g in this way is actually the entire group.

2) The group formed in this way, i.e. through successive powers of an element, is cyclic

3) The group is Abelian.
• May 30th 2010, 01:00 PM
roninpro
Yes, and you're done!
• May 30th 2010, 02:52 PM
GeoC
Thank you for the help along the way. (Bow)