For $n,i>0$ find a normalizer of $\langle \tau \sigma^i\rangle$ in $D_{2n }$ 
For $n,i>0$ find a normalizer of $\langle \tau \sigma^i\rangle$ in $D_{2n }$. I would like to know if my reasoning holds and I also block at one point. If someone could give a feedback and help (with developed answer), I would really appreciate it.

First, we notice that $H=\langle \tau \sigma^i\rangle=\{e,\tau\sigma^i\}$. Let $\sigma^k, \tau\sigma^k\in D_{2n}, k\ge 0$. If $\sigma^k, \tau\sigma^k\in N_{D_{2n}}(H)$, then:

*

*$\sigma^{-k}\tau\sigma^i\sigma^k=\tau\sigma^{2k+i}\in H$ (so equals to $e$ or $\tau\sigma^i)$  and


*$(\tau\sigma^k)^{-1}\tau\sigma^i\tau\sigma^k=\tau\sigma^{2k-i}\in H$ (so equals to $e$ or $\tau\sigma^i)$
But, in both cases we can't get $e$. So, both equalities must "give" us $\tau\sigma^i$.
As the order of $\sigma \in D_{2n}$ is equal to $n$, we have clearly that:
$2k+i \mod n=i \mod n$ for the first bullet point and $2k-i \mod n =i \mod n$ for the second.
We distinguish now 2 cases: $n$ even or $n$ odd.

*

*If $n$ is even, then :

$2k+i \mod n=i \mod n\iff 2k \mod n =0 \mod n \iff 2k=j\cdot n \implies k=\frac{j\cdot n}{2}$ and
$2k-i \mod n=i \mod n \iff 2k \mod n = 2i \mod n \iff 2k=j\cdot n+2i \implies k=\frac{j\cdot n}{2}+i$ for $j \in \mathbb{Z}$
Therefore $N_{D_{2n}}(H)=\{\sigma^{n/2}, \tau\sigma^{i+n/2},e,\tau\sigma^i\}$ for $n$ even

*

*If $n$ is odd, then:

$2k+i \mod n=i \mod n\iff 2k \mod n =0 \mod n \iff 2k=j\cdot n\implies k=0$
$\underbrace{2k-i \mod n=i \mod n \iff 2k \mod n = 2i \mod n \iff 2k=j\cdot n+2i \implies (?)}_{\text{for this case i can't conclude...} }$
I don't really see the "general" solution for $k$ in the last case. Probably, I should take $n=0$? I don't see other solutions and possibilities... If it is $n=0$ (but $n$ is not odd :( ), then the normalizer is $H$ itself.
So, $N_{D_{2n}}(H)=\{e,(?)\}$ if $n$ is odd.
 A: Remember that $0 \leq k < n$. You have two equations, depending on the case:
$2k + i \equiv i$ (mod n) $\iff 2k \equiv 0$ (mod n) (1)
$2k - i \equiv i$ (mod n) $\iff 2k \equiv 2i$ (mod n) (2)
Therefore, let's look at your solutions.
Case 1 - $n$ is even
In this case, let's analyse the prior possibilities.
From (1), we get $2k = jn$, where $j \in \{0, 1\}$ (since, if $j \geq2$, we would have $k \geq n$). Therefore, there are two valid possibilities: $k = 0$ and $k = \frac{n}{2}$.
Therefore if $\sigma^k \in N$, we must have one of the two $k$ above. You can check and see that they both work, so $e \in N$ and $\sigma^{\frac{n}{2}} \in N$.

From (2), we get $2k = jn + 2i \implies k = \frac{jn}{2} + i$.
If $i \geq \frac{n}{2}$, the solutions are $k = i$ and $(\frac{n}{2} + i) - n = i - \frac{n}{2}$. Otherwise, the two solutions are $k = i$ and $k = \frac{n}{2} + i$. The two pairs are effectively the same - I've just written them in reduced form modulo $n$.
Once again, we check and see that they work, meaning $\tau \sigma^i \in N$ and $\tau \sigma^{\frac{n}{2} + i} \in N$.
Therefore, $N = \{\tau \sigma^{\frac{n}{2} + i}, \sigma^{\frac{n}{2}}, e, \tau \sigma^i \}$
Case 2 - n is odd
Once again, let's analyse the equations.
From (1), $2k = jn$, with $j \in \{0, 1\}$. Since n is odd, $2k = n$ does not work, leaving us only with $2k = 0 \implies k = 0$.
So the only element of the kind $\sigma^k$ in $N$ is $e$.

Now we look at (2), $2k = jn + 2i$. For simplicity, suppose, at first, that $i \leq \lfloor{\frac{n}{2}}\rfloor$. The case $j = 0$ yields the solution $k = i$. And the case $j = 1$ does not yield a valid solution, because the right-hand side is odd.
Therefore, the only element of the form $\tau \sigma^k$ in $N$ is $\tau \sigma^i$.
Finally, if $i > \lfloor \frac{n}{2} \rfloor$, nothing is changed, as the solution for $j = 1$ does not exist.

From the discussion above, we conclude:

*

*If $n$ is even, the normalizer is $N = \{\tau \sigma^{\frac{n}{2} + i}, \sigma^{\frac{n}{2}}, e, \tau \sigma^i \} = H \cup \{\tau \sigma^{\frac{n}{2} + i}, \sigma^{\frac{n}{2}}\}$;

*If $n$ is odd, the normalizer is trivial, since it always contains the subgroup itself: $N = H$.

