What is $D_{10}/\left \langle \sigma \right \rangle$ where $D_{10}$ is the Dihedral Group or a regular Pentagon?

Question

$D_{10} = ({1, \sigma, \sigma^2, \sigma^3, \sigma^4, \tau, \sigma\tau, \sigma^2\tau, \sigma^3\tau, \sigma^4\tau}) $ where $\sigma = (12345)$ and $\tau = (13)(45)$.

I'm stuck on trying to calculate what $D_{10}/\left \langle \sigma \right \rangle$ is. I'm trying to find all the images of $f$ where $f: D_{10} \longrightarrow H $ with the map being a homomorphism.

Answer 1

$<\sigma>$ is a cyclic group of order $5$ so $D_{10}/<\sigma>$ is a group of order $2$. There's only one isomorphism class of groups of order $2$ and I'll leave you to determine what that is.

If you're really stuck you could construct the cosets.

Answer 2

There are a number of ways to approach this problem. One way is to observe any homomorphism $f: D_{10} \to H$ is completely determined by $f(\sigma)$ and $f(\tau)$. Note that since $f$ is a homomorphism, and:

$\sigma^5 = \tau^2 = 1$

that $f(\sigma)^5 = f(\tau)^2 = e_H$.

So the order of $f(\sigma)$ divides $5$, and the order of $f(\tau)$ divides $2$. But these are primes, so there's only so many possibilities.

Moreover, we must still have (because $f$ is a homomorphism and $\tau\sigma = \sigma^4\tau$) that: $f(\tau)f(\sigma) = f(\sigma)^4f(\tau)$.

So suppose the order of $f(\sigma) = 5$. If the order of $f(\sigma) = 1$ (that is $f$ maps $\tau$ to $e_H$), we have:

$f(\sigma) = f(\sigma)^4$, that is, cancelling; $e_H = f(1) = f(\sigma)^3$, contradicting that $f(\sigma)$ has order $5$.

So if $f(\sigma)$ has order $5$, then $f(\tau)$ must have order $2$, and $f(D_{10}) \cong D_{10}$ (we are using the fact that $\sigma,\tau$ generate $D_{10}$ subject to the relations:

$\sigma^5 = \tau^2 = 1,\ \tau\sigma = \sigma^4\tau$).

If $f(\sigma)$ has order $1$, and $f(\tau) = h \neq e_H$, then $f(D_{10})$ is the cyclic group of order $2$: $\{e_H,h\}$.

Otherwise, $f$ is trivial, and $f(D_{10}) = \{e_H\}$.

So there are just $3$ distinct isomorphism classes of homomorphic images of $D_{10}$: $D_{10}, C_2,C_1$. To each of these corresponds a unique normal subgroup (namely, $\text{ker }f$ in each case, respectively: $\{1\},\langle\sigma \rangle,D_{10}$).