Let $|G| = 105, K \lhd G$ and $|K| = 5$. Show that for every $x \in G, x^{21} \in K$.

Question

I'm learning abstract algebra, specifically quotient group and Lagrange's theorem, and need help to understand the solution to the following problem:

Let $|G| = 105, K \lhd G$ and $|K| = 5$. Show that for every $x \in G, x^{21} \in K$.

We use the symbols $|G|$ to denote the order of the group $G$ and $ K \lhd G$ to denote that $H$ is a normal subgroup of $G$.

Here's the solution (I have put question marks above the equalities/implications that I don't understand):

Since $G$ is finite, by Lagrange's theorem we have that

$$|G| = |G:K||K| \implies |G:K| = \frac{105}{5} = 21.$$

Because $H$ is a normal subgroup of $G$, $G/H$ becomes a group (quotient) under coset multiplication and $|G/K| = |G:K| = 21.$

$(Q1)$ If I understand this last equality correctly, the order of the quotient group $G/H$ is equal to the index of $K$ in $G$ which we found by Lagrange's theorem to be equal to $21$ (which is equal to the number of left cosets of $H$ in $G$). Is that correct?

Now for the part that I don't understand:

We have that $$(aK)^{21} \stackrel{?}{=} K \implies a^{21}K = K \stackrel{?}{\implies} a^{21} \in K.$$ Therefore, $\forall x \in G, x^{21} \in K$.

$(Q2)$ I don't understand why $(aK)^{21} = K$. I have written the quotient group in the form $G/K = \{aK : a \in G\}$ but it didn't help me much. The second implication is also unclear to me.

Answer 1

For Q1, you are correct (although you've written $H$ where you meant $K$ a few times).

For Q2, one way to approach this is to note that since $|K|$ and $|G/K|$ are relatively prime, we know that $K$ is the only subgroup of order $5$ (see here, for example). Clearly $\alpha^{105} = 1$, so that $(\alpha^{21})^5 = 1$. So either $\alpha^{21}=1\in K$ or else $\alpha^{21}$ has order $5$ in $G$. In this case, since $K$ is the only order 5 subgroup, we must have $\alpha^{21}\in K$.