Sylow counting - classifying groups of order 15

Question

let $G$ be a group of order $15$.

this is the argument I was given:

We have $15 = 3\times 5$ so we start with $p = 5$ We have by Sylow's theorem that $N_5 = 1 \mod 5$ so $N_5 = 1$ or $N_5 \geq 6$. Suppose that $N_5 \geq 6$ Then there exists at least six subgroups $K_i$ each of order $5$ then we note that $K_i \cong C_5$ and that $K_i \cup K_j = \{ 1 \}$ for $i \not= j$

Hence $\bigcup_{i=1}^6 K_i$ has $6 \times (5-1) = 24$ distinct elements of order $5$ but this contradicts that $|G| = 15 < 24$. Hence we have that $N_5 = 1$ i.e. there exists a unique subgroup $K$ of order $5$ with $K = C_5$. Now if $g \in G$ then gKg^{-1} is also a subgroup of order $5$ so $gKg^{-1} \subset K$ hence $K \lhd G$.

I will stop here because I have a question:

The lecturer stated that $gKg^{-1}$ is also a subgroup of order $5$. I don't understand why this is. I've manually proved it is a subgroup of order $5$ but, he stated it so freely so I feel as though there is an obvious deduction of why it is a subgroup of order $5$ - could someone tell me how it is?

many thanks

edit: sorry I have one more question, at the end of the proof I conclude that $G \cong C_5 \rtimes_h C_3$ with $h : C_3 \to \text{Aut} (C_5) \cong C_4$ but I don't know how to find $h$? I asked a friend he told me he thinks it's the trivial mapping $id(x) = x$ but he doesn't know why.

Answer 1

The conjugate of a subgroup $H$, i.e., $aHa^{-1}$ is always a subgroup, because it's closed under products.

What's its order? Well, the map $h \mapsto aha^{-1}$ is a bijection (this takes alittle proof, which I'll leave to you...hint: find the inverse function!) from $H$ to the subgroup, so it has the same order as $H$.

Answer 2

It is because conjugation is an isomorphism, hence is bijective.

You can show there is exactly one $K$-Sylow if you use the fact that, not only $ N_5\equiv 1\mod 5$, but also that $N_5 \mid 15/5$.

Added

For the second question, $h$ is a group homomorphism. But as $C_3$ has order $3$ and $\operatorname{Aut}(C_5)$ has order $4$, and the orders are coprime, the only homomorphism is the trivial one. Hence the semi-direct product is the direct product and $G$ is cyclic.

Note This is a general result on groups of order $pq\,$ ($p, q$ primes, $p<q$): if $p\nmid q-1$, they're cyclic.