Finding all elements of order 7 in a semi-direct product group

Question

Let $H=<h>,\ G=<g>,\ o(h)=7,\ o(g)=3$, and let $\alpha:G\rightarrow Aut(H)$ so that $\alpha(g)(h) = h^2$. Find all the elements of order 7 in $H\rtimes_\alpha G$.

We get that $\alpha(g^j)(h^i) = h^{i\cdot 2^j}$. So $(h^i, g^j)^k = (h^{i\cdot \sum\limits_{n=0}^{k-1} 2^{nj}}, g^{kj})$, and we want $7 \bigg| \sum\limits_{n=0}^{k-1} 2^{nj}$ and $3 \bigg| k$. So we can substitue $k=3m,\ \mathbb{Z} \ni m \geq 0$, and we need to find which m's and j's such that $m \in \mathbb{B},\ 0<j<3$ satisfy $7 \bigg| \sum\limits_{n=0}^{3m-1} 2^{nj}$. For m,j as such, every i should fit, so we justy multiply the result by 7.
I'm not sure how to find the j's and m's though...

Answer 1

As commented by darko, this is just Sylow's Theorems and we don't need to worry too much for the semidirect product (just a little...).

The group has order $\;21=7\cdot3\;$ , which means it has one unique Sylow $\;7\,-$ subgroup (which is then normal, by the way), and thus we already have $\;6\;$ elements of order $\;7\;$ in our group.

Since the given semidirect product is non-trivial (why?) it is then non-abelian (as the direct product of abelian groups is abelian itself), and thus there must be $\;7\;$ Sylow $\;3\,-\,$ subgroups (the other option is only one such subgroup, which would then be normal and then the semidirect product would be direct...), giving us $\;7\cdot2=14\;$ elements of order 3.

Thus, there are only $\;6\;$ elements of order $\;7\;$ .

Answer 2

Every nonunit element of $H$ (seen inside $H\rtimes G$) has order $7$.

In the other direction, note that $H$ is normal in $H\rtimes G$, and the quotient $(H\rtimes G)/H$ is isomorphic to $G$, and has order $|G|=3$.

If $x$ has order $7$ in $H\rtimes G$, then the image of $x$ in the quotient $(H\rtimes G)/H$ has order which divides $7$ and $3$, so it must be $1$, i.e., $x\in H$.