Problem
If $K \trianglelefteq G$ has prime index $p$, show that there exists $a \in G$ such that $G=K \cup K a \cup \cdots \cup K a^{p-1}$ is a disjoint union.
Attempt
Let $K \trianglelefteq G$. Suppose that $|G/K|=[G:K]=p$ where $p$ is prime. Hence, there is $a \in G$ such that $(Ka)^{p}=K$, namely $a^p=1$. Since $p$ is prime, $1$, $a$, $a^2$, $\cdots$, $a^{p-1}$ are different. Then, $K$, $Ka$, $\cdots$, $K a^{p-1}$ are different since $K$ is normal in $G$. Recalling $[G:K]=p$, we see that $$ G=K \cup K a \cup \cdots \cup K a^{p-1} $$ which is a disjoint union. $\blacksquare$
Question
- Is the proof valid?
- Is there a theorem to be mentioned for more rigorous proof?
Attempt after feedback
Let $K \trianglelefteq G$. Suppose that $|G/K|=[G:K]=p$ where $p$ is prime. Then $G/K$ is cyclic by the Lagrange's theorem. Then there is $Ka\in G/K$ such that $G / K= \langle Ka \rangle$. By the Lagrange's theorem, since $|G/K|=p$, we get $(Ka)^{p}=1_{G/K}$. Observe that $$ (Ka)^{p}=Ka^{p}=1_{G/K}=K $$ so that $a^{p}\in K$. Then, $K$, $Ka$, $\cdots$, $K a^{p-1}$ are different since $K$ is normal in $G$. Recalling $[G:K]=p$, we see that $$ G=K \cup K a \cup \cdots \cup K a^{p-1} $$ which is a disjoint union. $\blacksquare$
No, your proof is not correct. From $(Ka)^p=K$ you can't derive that $a^p=1$. As an example look at $\mathbb{Z}$ (and beware that we have an additive group here): If $p$ is any prime number, $K:=p\mathbb{Z}$ has index $p$. Now look at $K+1$: We have $p(K+1)=K$, but there is no element $a\in K+1$ s.t. $pa=1$, since $K+1$ is still a subset of the integers.
You can however adjust your proof by working on the quotient group $G/K$, which I think is what you intended to do and implicitly did, when you noted the existence of $a$. There we may actually do what you tried to do.