Functions
⎩ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎧ f 1 ( x ) = x f 2 ( x ) = 1 − x f 3 ( x ) = x 1 f 4 ( x ) = x − 1 x f 5 ( x ) = 1 − x 1 f 6 ( x ) = 1 − x 1
Here we define 6 functions as shown above.
( { f 1 , f 2 , . . . , f 6 } , ∘ ) , where ∘ being the function composition , defines a group.
This group is isomorphic to __________ .
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2 solutions
Moderator note:
For groups of small orders, due to the limited number of them, they can be easily classified by their actions on certain generators.
Wow! Your solution is more concise than mine.
The answer is S 3 .
First, there are only two different groups up to isomorphism, namely S 3 and Z 6 . Z 2 × Z 3 is isomorphic to Z 6 .
There are several ways to observe it is S 3 but not Z 6 .
First way is to check the power of each element. It is 1 , 2 , 2 , 2 , 3 , 3 respectively, which is same as S 3 . For Z 6 it should be 1 , 2 , 3 , 3 , 6 , 6 .
Second way is to identify f 1 as identity e , f 2 , f 3 , f 4 as the 3 2-cycles in S 3 , and f 5 , f 6 as the 2 3-cycles.
Another nice way to see what the group is actually doing is to see how it acts on 0 , 1 , ∞ . Then
f 1 is the identity
f 2 is the 2-cycle transposing 0,1
f 3 is the 2-cycle transposing 0, ∞
f 4 is the 2-cycle transposing 1, ∞
f 5 is the 3-cycle ( 0 ∞ 1 )
f 6 is the 3-cycle ( 0 1 ∞ )
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Great observation! This inspires me! With this observation it's obvious that it is S 3 .
Note that the product of cyclic groups Z 2 × Z 3 is isomorphic to the cyclic group Z 6 . For example, ( 1 , 1 ) is a generator.
More generally, for orders that are square-free, there is a unique abelian group of that order.
I have removed one of these options.
f 3 ( f 2 ( x ) ) = 1 − x 1 = f 6 ( x ) and f 2 ( f 3 ( x ) ) = 1 − x 1 = f 5 ( x ) . This group is isomorphic to S 3 , the only non-Abelian group of order 6.