Compute $H=\{\rho^3_0,\rho^3_1,\rho^3_2, \mu^3_0,\mu^3_1,\mu^3_2\}$ and show that $H$ is not a subgroup of $S_3$.

Question

Here is the question:

Suppose $G = S_3$. Compute $H=\{\rho^3_0,\rho^3_1,\rho^3_2, \mu^3_0,\mu^3_1,\mu^3_2\}$ and show that $H$ is not a subgroup of $S_3$.

I'm having trouble computing $H$. I know that the elements of $S_3$ are $\{(123),(132),(213),(231),(312),(321)\}$ so do I just pair these elements and say which they belong to of $H$?

For example, I have that

$$H = \{\rho^3_0 = (123), \rho^3_1 = (132), \rho^3_2 = (231), \mu^3_0= (321), \mu^3_1 = (231), \mu^3_2 = (312)\}. $$

And then I know to prove $H$ is a subgroup we need identity, closure, and inverse. However, I am not seeing how that would apply or how I would even start.

Answer 1

Using cycle notation for the non-identity elements and $e$ for the identity (see https://en.wikipedia.org/wiki/Permutation#Notations for details), $S_3 = \{e, (1, 2), (1, 3), (2, 3), (1, 2, 3), (1, 3, 2)\}$. Taking the third power of each element, we have:

$(1, 2)^3 = (1, 2)$

$(1, 3)^3 = (1, 3)$

$(2, 3)^3 = (2, 3)$

$e^3 = (1, 2, 3)^3 = (1, 3, 2)^3 = e$

So the set $H$ you’re interested in consists of 4 elements. By Lagrange’s theorem, it follows that $H$ cannot be a subgroup of $S_3$.