Show by specifying an isomorphism that the group of sixth unit roots $\mu_6$ is isomorphic to the subgroup

Question

Show by specifying an isomorphism that the group of sixth unit roots $\mu_6$ is isomorphic to the subgroup $\langle (1 \enspace 3 \enspace 5), (2 \enspace 4) \rangle$ of $S_5$.

Attempt:

Show: Group $\mu_6$ isomorphic to $\langle (1 \enspace 3 \enspace 5), (2 \enspace 4) \rangle$ of $S_5$

Let H = $\langle (1 \enspace 3 \enspace 5) (2 \enspace 4) \rangle$.

The element $\sigma = (1 \enspace 3 \enspace 5) (2 \enspace 4)$ has the order: $$Ord(\sigma) = kgV(3,2)=6 $$

so H is cyclic of order 6.

By script, a group G is isomorphic to H if and only if G is also cyclic to order 6.

The group $\mu_6$ is cyclic of order 6, i.e. $\mu_6$ is isomorphic to H.

(But what is meant by "show by specifying an isomorphism"? Can someone also check my reasoning above?)

Edit:

The only information I have given so far on how to set up the isomorphism would be this one from our script:

Example 1.3.6 Let $ G=\langle g\rangle $ and $ H=\langle h\rangle $ be two cyclic groups of order $ n $. The mapping $$ \varphi: G \rightarrow H, \quad g^{i} \mapsto h^{i} . $$ is a bijective group isomorphism. So $ G $ and $ H $ are isomorphic.

To do this, I know I need to construct a bijection that preserves products as in the definition of an isomorphism. I’m at a loss for how to construct such a bijection, because I don’t see how any function I can think of can be both injective and surjective. Any guidance?

Answer 1

If we set $\omega:= e^{2i \pi/3}$, any $6$th root of unity can be given the form:

$$s \omega^k, \ \ \underbrace{s=\pm 1}_{\text{accounts for the (2 4) part}}, \ \underbrace{k=0,1,2}_{\text{accounts for the (1 3 5) part}}$$

providing an isomorphism in a natural way.

Answer 2

To "show by specifying an isomorphism" means to write down an explicit formula for the isomorphism.

Example 1.3.6 contains such a formula already, namely $\varphi(g^i)=h^i$, although part of your job is to say what $g$ and $h$ are in the context of the problem you were given (which you have almost already done). Also, if one is working through that exercise, then another job is to prove that this formula is well-defined, is a homomorphism, is surjective, and is injective.

Answer 3

$$1\color{red}23\color{red}45\mapsto 1$$ $$3\color{red}45\color{red}21\mapsto e^{\frac{2i\pi}{6}}$$ $$5\color{red}21\color{red}43\mapsto e^{\frac{4i\pi}{6}}$$ $$1\color{red}43\color{red}25\mapsto -1$$ $$3\color{red}25\color{red}41\mapsto e^{\frac{8i\pi}{6}}$$ $$5\color{red}41\color{red}23\mapsto e^{\frac{10i\pi}{6}}$$ $$1\color{red}23\color{red}45\mapsto 1$$

Note : Instead of writing $\begin{pmatrix}1 & 2 & 3 & 4 & 5 \\3 & 4 & 5 & 2 & 1\end{pmatrix}$, we simply wrote $34521$.