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I didn't understand how to write down the alternating group A3. Is this the group consisting of only the even permutations? Also, what familiar group is this isomorphic to?

Anastasia
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    Yes, this is the group of only the even permutations of ${1,2,3}$. For the second question, start by writing down this group; it will have a very small order, so it should be easily recognizable from its Cayley table. – Nick Peterson Mar 16 '17 at 14:57
  • "Also, what familiar group is this isomorphic to?" It depends which groups you are familiar with. I mean, one could presume that you are familiar with some groups, but that same person might also presume that you are familiar with other kinds of early results of group theory which, however, you do not seem keen to use. –  Mar 16 '17 at 14:58

3 Answers3

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Yes, $A_3$ is the set of all even permutations in $S_3 = \{id, (12), (13), (23), (123), (132)\}$.

Remember that an even permutation can be written as the product of an even number of transpositions. The identity of any symmetric group is even, because id can be written as the product of two transposition. In this case, e.g. $id = (1,2)(2, 1)$.

Note also that $(123) = (12)(23)$, and $(132) = (13)(32).$ So, $$A_3 = \{id, (123), (132)\}.$$

Since $|A_3| = 3$, and the fact that there is only one group, up to isomorphism, of order $3$, $$A_3 \cong \mathbb Z_3,$$ where $\mathbb Z_3$ is the cyclic group under addition modulo $3$.

amWhy
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  • You can also write the identity as the product of no transpositions. Seems a more reasonable way to write down the identity in terms of transpositions, and is in fact (obviously) the minimal number of elementary transpositions one needs. – Pedro Mar 16 '17 at 16:39
  • True...@Pedro. Thanks for chiming in! – amWhy Mar 16 '17 at 16:58
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You can also have an isomorphism with the following matrix group:

$$I=\begin{pmatrix}1&0&0\\0&1&0\\0&0&1\end{pmatrix}, \ \ C=\begin{pmatrix}0&0&1\\1&0&0\\0&1&0\end{pmatrix}, \ \ C^2=\begin{pmatrix}0&1&0\\0&0&1\\1&0&0\end{pmatrix}$$

Explanation: if $e_1,e_2,e_3$ is the canonical basis

  • $I$ will sent all $e_i$ on themselves,

  • $C$ will send $e_1$ onto $e_2$, $e_2$ onto $e_3$, $e_3$ onto $e_1$, that's a cycle !

  • $C^2$ will... be as well $C^{-1}$, because $C^3=I.$

Jean Marie
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Its elements are the $3$-cycles and the identity

$\begin{cases} f(1)=2, f(2)=3, f(3)=1 \\ g(1)=3, g(2)=1, g(3)=2 \\ Id \end{cases}$.

zwim
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Tsemo Aristide
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