[SOLVED] Groups of 3,4,5 elements abelian

Oct 2009
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This is pretty straight forward for groups of 3 and 4 elements. Is there a better approach to groups with 5 elements?
 
Nov 2009
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Are there any other tools that you are allowed to use? Langrange's Theorem?
 
Oct 2009
195
19
Are there any other tools that you are allowed to use? Langrange's Theorem?
It's the first chapter introducing Groups. Here's what I've got so far:

  • Basic Axioms of Groups
  • Unique Identity
  • Unique Inverses
  • \(\displaystyle (a^{-1})^{-1} = a\)
  • \(\displaystyle (a*b)^{-1} = b^{-1}*a^{-1}\)
  • Cancellation
 
Dec 2009
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I was wondering, Can you use that the order of all group elements must divide the groups order? Otherwise it's trivial since 5 is prime, hence G is cyclic (can be generated by one element).
 
Nov 2009
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I think this will work.

Let \(\displaystyle a,b \in G\).

If \(\displaystyle a=b,\) then \(\displaystyle ab=ba\).
If \(\displaystyle a=b^{-1},\) then \(\displaystyle ab=ba=e\).

So suppose that the two aren't equal and inverses of each other. Then \(\displaystyle ab\) is an element that is not equal to \(\displaystyle e,a,\) or \(\displaystyle b\). This goes for \(\displaystyle ba\) as well.

Now suppose that \(\displaystyle ab \not= ba\). Then we have

\(\displaystyle G=\{e,a,b,ab,ba \},\)

which means that \(\displaystyle a^{-1},b^{-1} \in \{ab,ba\}.\) If \(\displaystyle a^{-1}=ab\) and \(\displaystyle b^{-1}=ba,\) then

\(\displaystyle a^2b=e \Longrightarrow a^2=ba \Longrightarrow a=b,\)

which is a contradiction. The case for \(\displaystyle a^{-1}=ba, \, b^{-1}=ab\) is similar.
 
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Oct 2009
195
19
I was wondering, Can you use that the order of all group elements must divide the groups order? Otherwise it's trivial since 5 is prime, hence G is cyclic (can be generated by one element).
I don't think so. The book has defined a cyclic group, but not introduced any properties regarding cyclic groups or the order of groups in general.

I'm thinking maybe they want me to build more operation tables. But that gets much trickier for groups of 5 elements.
 
Oct 2009
195
19
I think this will work.

Let \(\displaystyle a,b \in G\).

If \(\displaystyle a=b,\) then \(\displaystyle ab=ba\).
If \(\displaystyle a=b^{-1},\) then \(\displaystyle ab=ba=e\).

So suppose that the two aren't equal and inverses of each other. Then \(\displaystyle ab\) is an element that is not equal to \(\displaystyle e,a,\) or \(\displaystyle b\). This goes for \(\displaystyle ba\) as well.

Now suppose that \(\displaystyle ab \not= ba\). Then we have

\(\displaystyle G=\{e,a,b,ab,ba \},\)

which means that \(\displaystyle a^{-1},b^{-1} \in \{ab,ba\}.\) If \(\displaystyle a^{-1}=ab\) and \(\displaystyle b^{-1}=ba,\) then

\(\displaystyle a^2b=e \Longrightarrow a^2=ba \Longrightarrow a=b,\)

which is a contradiction. The case for \(\displaystyle a^{-1}=ba, \, b^{-1}=ab\) is similar.
I thought there was a way to do it using just the properties. I was hoping someone could confirm that, but you did the whole thing. Thanks :)