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Find an abelian subgroup of A5 of order 4
- Thread starter tegra97
- Start date
(2345)(2354) = (243) (doing composiion in the usual way), so No, it is not a group.
Also All elements have to be products of 2-cycles.
What about {(1,2)(3,4)(5), (1,3)(2,4)(5), (2,3)(4,1)(5), (1)(2)(3)(4)(5)}. Each is its own inverse and the product of any distinct ones is the other. To show its abelian, just write out the products.
Also All elements have to be products of 2-cycles.
What about {(1,2)(3,4)(5), (1,3)(2,4)(5), (2,3)(4,1)(5), (1)(2)(3)(4)(5)}. Each is its own inverse and the product of any distinct ones is the other. To show its abelian, just write out the products.
The order of an individual ELEMENT is equal to the LCM of the orders of the cycles of which it is made up.
For example if \(\displaystyle \L \sigma \in S_n\) and \(\displaystyle \L \sigma = (k_1-cycle)(k_2-cycle)\cdot\cdot\cdot(k_n-cycle)\) where the cycles are disjoint then we have that \(\displaystyle \L | \sigma | = LCM(k_1,k_2,...,k_n).\)
For example if \(\displaystyle \L \sigma \in S_n\) and \(\displaystyle \L \sigma = (k_1-cycle)(k_2-cycle)\cdot\cdot\cdot(k_n-cycle)\) where the cycles are disjoint then we have that \(\displaystyle \L | \sigma | = LCM(k_1,k_2,...,k_n).\)
Do you know how to multiply (i.e. compose) two permutations?If not, consult your textbook. What is needed to show that some set with an operation is a group? It must have an identity, an inverse and it must be CLOSED (Associativity is free since A_n is associative). The first two are trivial.
In your case, given \(\displaystyle \sigma, \,\, \tau \,\, \in \,\, S=\{(12)(34)(5),\,\,(13)(24)(5),\,\,(23)(41)(5),\,\,(1)(2)(3)(4)(5)\}\) you need to show \(\displaystyle \sigma \tau \,\, \in S\). Also that \(\displaystyle \sigma \tau = \tau \sigma\) to show commutativity.
Generally closure tests (resp. abelian) are done arbitrarily. However, given this specific set closure should be done for each specific element (unless you see some general rule that I don't). A simple operation table (called a "Cayley Table") will show this clearly.
In your case, given \(\displaystyle \sigma, \,\, \tau \,\, \in \,\, S=\{(12)(34)(5),\,\,(13)(24)(5),\,\,(23)(41)(5),\,\,(1)(2)(3)(4)(5)\}\) you need to show \(\displaystyle \sigma \tau \,\, \in S\). Also that \(\displaystyle \sigma \tau = \tau \sigma\) to show commutativity.
Generally closure tests (resp. abelian) are done arbitrarily. However, given this specific set closure should be done for each specific element (unless you see some general rule that I don't). A simple operation table (called a "Cayley Table") will show this clearly.
Me too.tegra97 said:I think I'm making this problem a lot harder then it actually is. :?
What does the composition (12)(34)(5) OF (13)(24)(5) equal? Is it also in the set? Does it equal (13)(24)(5) OF (12)(34)(5)? If the answers to these questions are "YES" for every two elements in the set then the set is closed and abelian. Note that you can ommit 1-cycles like (5) since all it does is fix 5, and the identity (1)(2)(3)(4)(5) can be written as (1).
The hard part is finding the subset. Checking that the subset is an abelian subgroup should be the easy part.