Abelian group
- Tolaso J Kos
- Administrator
- Posts: 867
- Joined: Sat Nov 07, 2015 6:12 pm
- Location: Larisa
- Contact:
Abelian group
Prove that a group $(G, *)$ that has either $1$ or $2$ or $3$ or $4$ elements is abelian.
Imagination is much more important than knowledge.
-
- Community Team
- Posts: 426
- Joined: Mon Nov 09, 2015 1:52 pm
Re: Abelian group
Hi Tolaso.
Let \(\displaystyle{\left(G,\cdot\right)}\) be a group.
If \(\displaystyle{|G|=1}\), then \(\displaystyle{G=\left\{e\right\}}\) and it's ok.
If \(\displaystyle{|G|=2\,\,,G=\left\{e,a\right\}}\), then \(\displaystyle{\left(G,\cdot\right)}\) is abelian
and \(\displaystyle{\left(G,\cdot\right)\simeq \left(\mathbb{Z}_{2},+\right)}\) .
Let \(\displaystyle{|G|=3\,\,,G=\left\{e,a,b\right\}}\).
1st proof
We have that \(\displaystyle{e\,a=a\,e=a\,\,,e\,b=b\,e}\). If \(\displaystyle{a\,b=a}\) or \(\displaystyle{a\,b=b}\)
then \(\displaystyle{b=e}\) or \(\displaystyle{a=e}\), a contradiction. So,
\(\displaystyle{a\,b=e=b\,a}\) and \(\displaystyle{\left(G,\cdot\right)}\) is abelian.
2nd proof
If \(\displaystyle{H}\) is a subgroup of \(\displaystyle{\left(G,\cdot\right)}\), then
\(\displaystyle{|H|\mid |G|=3\implies |H|\in\left\{1,3\right\}\implies H\in\left\{\left\{e\right\},G\right\}}\).
So, \(\displaystyle{\forall\,x\in G-\left\{e\right\}, G=\langle{x\rangle}}\) and \(\displaystyle{\left(G,\cdot\right)}\)
is abelian as cyclic and thus \(\displaystyle{\left(G,\cdot\right)\simeq \left(\mathbb{Z}_{3},+\right)}\) .
Last case
Let \(\displaystyle{|G|=4=2^2}\). If \(\displaystyle{x\,,y\in G}\), then
\(\displaystyle{|\langle{x\rangle}|\,,|\langle{y\rangle}|\in\left\{1,2,4\right\}}\) .
If \(\displaystyle{|\langle{x\rangle}|=1}\) or \(\displaystyle{|\langle{y\rangle}|=1}\), then
\(\displaystyle{x=e}\) or \(\displaystyle{y=e}\) and \(\displaystyle{x\,y=y\,x\in\left\{x,y\right\}}\) .
If \(\displaystyle{|\langle{x\rangle}|=4}\) or \(\displaystyle{|\langle{y\rangle}|=4}\) , then
\(\displaystyle{G=\langle{x\rangle}}\), which means that \(\displaystyle{G}\) is abelian and
\(\displaystyle{\left(G,\cdot\right)\simeq \left(\mathbb{Z}_{4},+\right)}\) .
Let \(\displaystyle{|\langle{x\rangle}|=2=|\langle{y\rangle}|}\) . Then,
\(\displaystyle{\langle{x\rangle}\cap \langle{y\rangle}\leq \langle{x\rangle}}\), so, if
\(\displaystyle{\langle{x\rangle}\cap \langle{y\rangle}=\langle{x\rangle}\iff \langle{x\rangle}\subseteq \langle{y\rangle}}\)
and then \(\displaystyle{x\,y=y\,x}\). If \(\displaystyle{\langle{x\rangle}\cap \langle{y\rangle}=\left\{e\right\}}\), then
since \(\displaystyle{\langle{x\rangle}\,,\langle{y\rangle}}\) are normal subgroups, we get
\(\displaystyle{G=\langle{x\rangle}\cdot \langle{y\rangle}\simeq \mathbb{Z}_{2}\times \mathbb{Z}_{2}}\) .
Note
The usual way is to make the board of multiplication. However, the above solution, gives us also
all the groups of order \(\displaystyle{1,2,3}\) and \(\displaystyle{4}\) .
Let \(\displaystyle{\left(G,\cdot\right)}\) be a group.
If \(\displaystyle{|G|=1}\), then \(\displaystyle{G=\left\{e\right\}}\) and it's ok.
If \(\displaystyle{|G|=2\,\,,G=\left\{e,a\right\}}\), then \(\displaystyle{\left(G,\cdot\right)}\) is abelian
and \(\displaystyle{\left(G,\cdot\right)\simeq \left(\mathbb{Z}_{2},+\right)}\) .
Let \(\displaystyle{|G|=3\,\,,G=\left\{e,a,b\right\}}\).
1st proof
We have that \(\displaystyle{e\,a=a\,e=a\,\,,e\,b=b\,e}\). If \(\displaystyle{a\,b=a}\) or \(\displaystyle{a\,b=b}\)
then \(\displaystyle{b=e}\) or \(\displaystyle{a=e}\), a contradiction. So,
\(\displaystyle{a\,b=e=b\,a}\) and \(\displaystyle{\left(G,\cdot\right)}\) is abelian.
2nd proof
If \(\displaystyle{H}\) is a subgroup of \(\displaystyle{\left(G,\cdot\right)}\), then
\(\displaystyle{|H|\mid |G|=3\implies |H|\in\left\{1,3\right\}\implies H\in\left\{\left\{e\right\},G\right\}}\).
So, \(\displaystyle{\forall\,x\in G-\left\{e\right\}, G=\langle{x\rangle}}\) and \(\displaystyle{\left(G,\cdot\right)}\)
is abelian as cyclic and thus \(\displaystyle{\left(G,\cdot\right)\simeq \left(\mathbb{Z}_{3},+\right)}\) .
Last case
Let \(\displaystyle{|G|=4=2^2}\). If \(\displaystyle{x\,,y\in G}\), then
\(\displaystyle{|\langle{x\rangle}|\,,|\langle{y\rangle}|\in\left\{1,2,4\right\}}\) .
If \(\displaystyle{|\langle{x\rangle}|=1}\) or \(\displaystyle{|\langle{y\rangle}|=1}\), then
\(\displaystyle{x=e}\) or \(\displaystyle{y=e}\) and \(\displaystyle{x\,y=y\,x\in\left\{x,y\right\}}\) .
If \(\displaystyle{|\langle{x\rangle}|=4}\) or \(\displaystyle{|\langle{y\rangle}|=4}\) , then
\(\displaystyle{G=\langle{x\rangle}}\), which means that \(\displaystyle{G}\) is abelian and
\(\displaystyle{\left(G,\cdot\right)\simeq \left(\mathbb{Z}_{4},+\right)}\) .
Let \(\displaystyle{|\langle{x\rangle}|=2=|\langle{y\rangle}|}\) . Then,
\(\displaystyle{\langle{x\rangle}\cap \langle{y\rangle}\leq \langle{x\rangle}}\), so, if
\(\displaystyle{\langle{x\rangle}\cap \langle{y\rangle}=\langle{x\rangle}\iff \langle{x\rangle}\subseteq \langle{y\rangle}}\)
and then \(\displaystyle{x\,y=y\,x}\). If \(\displaystyle{\langle{x\rangle}\cap \langle{y\rangle}=\left\{e\right\}}\), then
since \(\displaystyle{\langle{x\rangle}\,,\langle{y\rangle}}\) are normal subgroups, we get
\(\displaystyle{G=\langle{x\rangle}\cdot \langle{y\rangle}\simeq \mathbb{Z}_{2}\times \mathbb{Z}_{2}}\) .
Note
The usual way is to make the board of multiplication. However, the above solution, gives us also
all the groups of order \(\displaystyle{1,2,3}\) and \(\displaystyle{4}\) .
Create an account or sign in to join the discussion
You need to be a member in order to post a reply
Create an account
Not a member? register to join our community
Members can start their own topics & subscribe to topics
It’s free and only takes a minute
Sign in
Who is online
Users browsing this forum: No registered users and 14 guests