let $H$ be a subgroup of a finite group $G$. Assume that $G$ is odd and $|G|=3|H|$ Prove that $H$ is normal in $G$
An hint is given for the solution as follows: let $\ a \in G\ s.t. a \not\in H$
Define $\phi:H\rightarrow \mathbb{Z}_2\ s.t.\ \phi(h)=0$ if $\ haH=aH$, $\phi(h)=1$ if $\ haH=a^2H$
Then $\phi(H)=\{0\}$ so $\phi(H)=\{0\}=Ker(\phi)$ so $H$ is normal.
My misunderstanding: Previously I had shown that if $a \not\in H$, there are 3 unique cosets of $H$ in $G$ ($H,aH,a^2H$). So if $h \in H$, and $\phi(h)=0$
$$(*)\implies haH=aH\implies h\in aH \implies H\cap aH\neq \emptyset\implies aH=H$$
Which is obviously a contradiction. Could help me understand
Why is statement $(*)$ not true? (Edit: this has been made clear to me now)
How would I show that $\phi(h)=0\ \forall h \in H$ ?
Assuming you have verified that $\phi$ is a homomorphism, recall that $H / \ker(\phi) \cong \text{im}(\phi)$, so $|H| / |\ker(\phi)| = |\text{im}(\phi)|$. Now $|G|$ is odd, so $|H|$ is odd, so $|H| / |\ker(\phi)|$ is odd, which means that $|\text{im}(\phi)|$ is odd. Since $\text{im}(\phi)$ is a subgroup of $\mathbb Z_2$, its order must be $1$ or $2$, and of these possibilities, only $1$ is odd. Therefore $|\text{im}(\phi)| = 1$, so $|\ker(\phi)| = |H|$, so $\ker(\phi) = H$.