Deduce that $H$ has no elements of finite order other than the identity element. Given operation $*$ defined on the set $G$, where $G =\{(a, b)\mid a, b \in \mathbb{Q}\}$ with $a$ and $b$ not both
zero, $(a, b) * (c, d) = (ac + 3bd, ad + bc)$.
Prove that subset $H = \{(a, 0)\mid a \in \mathbb{Q} \land a\neq 0\}$ is a subgroup of $G$. Find $(a, 0)^r$
for $r \in \mathbb{Z^+}$, where $(a, 0) \in H$ and deduce that $H$ has no elements of finite order other than the
identity element.
Attempt
I found identity element of $G$ as $(1,0)$ to prove $H$ is a subgroup of $G$ I took $(a,0),(b,0)\in H$ then I proved $(a,0)*(b,0)^{-1} \in H$ so $H$ is a subgroup of $G$
Next part, $(a, 0)^r=(a,0)*(a,0)*\dots*(a,0)=(a^r,0)$ so then,
$$(a, 0)^r=(a^r,0)=(1, 0).$$
This implies $a^r=1$ so $a$ can be $-1$ or $1$
If $a=1$ then $(1,0)$ we can neglect it since it is an identity element so we have another element $(-1,0) $ it is also have a finite order but the question is said to deduce there is no elements of finite order other than the identity element.
Is there anything wrong my steps?
Thank you!
 A: You're correct.
It is a subgroup.
I will use the one-step subgroup test. (Note that I have to show $\varnothing\neq H\subseteq G$.)
Observe that $1\in \Bbb Q$ and $1\neq 0$, so $(1,0)\in H$. Hence $H\neq\varnothing$.
Let $(a,0)\in H$. Then $a\neq 0$, so $a$ and $0$ are not both zero but are both rational. Hence $(a,0)\in G$. Hence $H\subseteq G$.
Let $A=(a,0), B=(b,0)\in H$. Then
$$\begin{align}
AB^{-1}&=(a,0)*(b,0)^{-1}\\
&=(a,0)*(b^{-1}, 0)\\
&=(ab^{-1}+3(0)(0), a(0)+(0)b^{-1})\\
&=(ab^{-1}, 0),
\end{align}$$
which is in $H$ since $ab^{-1}\in \Bbb Q\setminus \{0\}$ as $a,b\in\Bbb Q\setminus \{0\}$.
Thus $(H,*)\le (G,*)$.
The element $(-1,0)\in H$ has order two.
Indeed, if $r\in \Bbb Z^+$, then
$$\begin{align}
(a,0)^r&=\underbrace{(a,0)*\dots*(a,0)}_{r\text{ times}}\\
&=(a^r, 0),
\end{align}$$
since the second argument of $(a,0)$ is zero, meaning $3(0)(0)=a(0)=(0)a=0$.
But $a\neq 0$ by definition of $H$, so, since also $a\in \Bbb Q$, if
$$(a,0)^r=(a^r,0)=e_G=(1,0),$$
then $a^r=1$, meaning $a=\pm 1$ if $\pm r>0$, depending on whether $r$ is odd or even.
This can be checked directly.
Indeed,
$$\begin{align}
(-1,0)^2&=(-1,0)*(-1,0)\\
&=((-1)(-1)+3(0)(0), -1(0)+0(-1))\\
&=(1,0)
\end{align}$$
and $(-1,0)\in H$.
