In the book Contemporary Abstract Algebra by Gallian. Chapter 3 (edition 9). Exercise 3 says
What can you say about a subgroup of $D_3$ that contains $R_{240}$ and a reflection $F$? What can you say about a subgroup of $D_3$ that contains two reflections.
My attempt: The smallest subgroup of $D_3$ that contains $R_{240}$ and $F$ is $ ⟨R_{240},F ⟩$, which is equal to $D_3$. Thus a subgroup of $D_3$ that contains $R_{240}$ and $F$ is the group itself.
Now for the latter part. Suppose $F$ and $F'$ be two reflections of $D_3$. Using similar reasoning as above we see $ ⟨F,F' ⟩=${$F,F',F^2,F'^2,FF',F'F$}$=${$R_0,R_{120},R_{240},F,F'$}. But this is not group!! As closure fails because third reflection is not in the set.
Question\
- Where is the mistake in my above proof?\
- How to generate subgroup by two reflections?\
- $ ⟨F,F' ⟩=${$F,F',F^2,F'^2,FF',F'F$}$=${$R_0,R_{120},R_{240},F,F'$} is this correct expression for $ ⟨F,F' ⟩$ because i am confused how to generate subgroup for more than one element.
The dihedral group $D_3 = \{1,R_{120},R_{240},S,R_{120}S,R_{240}S\}$ contains three reflections, namely $S$, $R_{120}S$ and $R_{240}S$. Three cases are thus possible :
1. $F = S$ and $F' = R_{120}S$
In that case, a rotation can be reconstructed as follows : $R_{120} = F'F$. In consequence, we end up with your first situation, i.e. $\langle F,F' \rangle = \langle R_{120},S \rangle = D_3$.
2. $F = S$ and $F' = R_{240}S$
It works exactly the same way as the previous case.
3. $F = R_{120}S$ and $F' = R_{240}S$
The rotation $R_{240}$ is recovered by $FF' = R_{120}SR_{240}S = R_{120}R_{120} = R_{240}$, while the simple reflection is generated by $FF'F = R_{240}F = S$, so that we end up again with the first situation, where $\langle F,F' \rangle = \langle R_{240},S \rangle = D_3$.