# Theorem 9.7 Classification of Groups or order p^(2)

• Jan 20th 2011, 12:48 PM
santiagos11
Theorem 9.7 Classification of Groups or order p^(2)
This theorem is taken from Contemporary Abstract Algebra, Ed. 7.

If G is a group and |G|=p^(2) where p is a prime, then G is isomorphic to Z(p^2) or
Z(p) + Z(p).

Note: Here Z(n) is the group under addition modulo n. Also, if G and H are groups, then G + H represents the external direct product of G and H.

Suppose G is not isomorphic to Z(p^2). Then G does not have an element of order p^2. For if G had an element of order p^2, then G is isomorphic to Z(p^2), a contradiction. (I omit some details here since I understand it). Thus by Lagrange Theorem, every non-identity element of G must have order p.

(The next claim, I am not able to understand)

Claim: for any a in G, the subgroup <a> is normal in G

Note: <a> means the group/subgroup generated by a

I am very close to undertanding the book's proof, but there are some missing details that I do not seem to be able grasp.
• Jan 20th 2011, 01:26 PM
roninpro
There are many ways to prove this fact, so it would be easier to tailor something for you if we knew your background. Could you elaborate on some of the other material you have covered? In particular, one proof that I know involves using symmetry groups and group action - are you familiar with them?
• Jan 20th 2011, 01:50 PM
Drexel28
Quote:

Originally Posted by santiagos11
This theorem is taken from Contemporary Abstract Algebra, Ed. 7.

If G is a group and |G|=p^(2) where p is a prime, then G is isomorphic to Z(p^2) or
Z(p) + Z(p).

Note: Here Z(n) is the group under addition modulo n. Also, if G and H are groups, then G + H represents the external direct product of G and H.

Suppose G is not isomorphic to Z(p^2). Then G does not have an element of order p^2. For if G had an element of order p^2, then G is isomorphic to Z(p^2), a contradiction. (I omit some details here since I understand it). Thus by Lagrange Theorem, every non-identity element of G must have order p.

(The next claim, I am not able to understand)

Claim: for any a in G, the subgroup <a> is normal in G

Note: <a> means the group/subgroup generated by a

I am very close to undertanding the book's proof, but there are some missing details that I do not seem to be able grasp.

We don't know the book's proof, so how can we help you understand it? One way to prove this is to note by the class equation that every $p$-group has non-trivial center. So, if $|G|=p^2$ then $\left|\mathcal{Z}\left(G\right)\right|=p,p^2$. If $\left|\mathcal{Z}\left(G\right)\right|=p$ then $\left|G/\mathbb{Z}\left(G\right)\right|=p$ and so $G/\mathcal{Z}\left(G\right)$ is cyclic. But, a group $G$ is abelian if and only if $G/\mathcal{Z}\left(G\right)$ is cyclic, and thus we may conclude that $\mathcal{Z}\left(G\right)=G$ which contradicts that $\left|\mathcal{Z}\left(G\right|\right|=p$. We may therefore conclude that $\left|\mathcal{Z}\left(G\right)\right|=p^2$ and thus $\mathcal{Z}\left(G\right)=G$. It follows that every group of order $p^2$ is abelian. The rest follows from the structure theorem for finitely generated abelian groups where we clearly have that $G\cong C_{p^2}$ if there exists an element of order $p^2$ and $G\cong C_p\oplus C_p$ if all elements are of order $p$.
• Jan 20th 2011, 04:42 PM
santiagos11
Books proof:
...for any $a$ in $G$, the subgroup $$ is normal in $G$. If this is not the case then there is an element $b$ in $G$ such that $bab^{-1}$ is not in $$...

My interpretation:
For any a in G, the subgroup <a> is normal in G. Suppose not. Then there exists some a in G such that <a> is not normal in G, which is equivalent to the following statement: There exists some a and b in G such that b<a> != <a>b. But this is true if and only if $ \neq\ bb^{-1}$. Thus my question is: From these facts how can I deduce that $bab^{-1}$ is not in $$
• Jan 20th 2011, 04:46 PM
Drexel28
Quote:

Originally Posted by santiagos11
Books proof:
...for any $a$ in $G$, the subgroup $$ is normal in $G$. If this is not the case then there is an element $b$ in $G$ such that $bab^{-1}$ is not in $$...

My interpretation:
For any a in G, the subgroup <a> is normal in G. Suppose not. Then there exists some a in G such that <a> is not normal in G, which is equivalent to the following statement: There exists some a and b in G such that b<a> != <a>b. But this is true if and only if $ \neq\ bb^{-1}$. Thus my question is: From these facts how can I deduce that $bab^{-1}$ is not in $$

Think about it this way. If $bab^{-1}\in\langle a\rangle$ then $ba^kb^{-1}=\left(bab^{-1}\right)^k\in\langle a\rangle$ for all $k\in\mathbb{Z}$. In other words $b\langle a\rangle b^{-1}\subset\langle a\rangle$ but since $\#\left(b\langle a\rangle b^{-1}\right)=|\langle a\rangle|$ this would imply that $b\langle a\rangle b^{-1}=\langle a\rangle$ contradictory to assumption.
• Jan 20th 2011, 06:18 PM
santiagos11
Thank you, I got the rest of the proof. Sometimes, as math becomes more complex, so does our thinking, and thus we do not see the easy things.
• Jan 20th 2011, 06:37 PM
Drexel28
Quote:

Originally Posted by santiagos11
Thank you, I got the rest of the proof. Sometimes, as math becomes more complex, so does our thinking, and thus we do not see the easy things.

Good job! I agree!