Let $G=GL(2,\mathbb R)$ and $ H =\left\{
\left[\begin{array}{ccc|c}
a & 0 \\
0 & b
\end{array} \right]:\mbox {$a$ and $b$ are nonzero integers }\right \}$ under the operation matrix multiplication. Disprove that $H$ is a subgroup of $G=GL(2,\mathbb R)$.
Well I have learned that I have to prove that:
1) Show that e∈H (where e is the identity)
2) Assume that a∈H , b∈H
3) Show that ab∈H
4) Show that $(ab)^{-1}$ (Inverse)
So I know that It does not hold, But how do I prove that?
When I try to prove that e∈H
I only get the Identity matrix and that holds because I get that a
and b
are nonzero integers.
When I prove that a and b is in the set I get that it is because both a and b are nonzero integers and that is the identity matrix.
Now I proved that a.b∈H as follows:
$$ H =
\left[ \begin{array}{ccc|c}
a & 0 \\
0 & b
\end{array} \right]
\left[ \begin{array}{ccc|c}
c & 0 \\
0 & d
\end{array} \right] =
\left[ \begin{array}{ccc|c}
ac & 0 \\
0 & bd
\end{array} \right]
$$
With a = b = c = d = 1
I get the Identity matrix again.
Now I know if a = b = 2
the subgroup will not hold because the inverse will be a set of rational numbers and H is only a subgroup if it contains only Integer.
Is my reasoning correct and if not where did I go wrong?
Best Answer
You're fourth step is a bit of an overkill, in terms of testing whether a subset is a subgroup. Indeed, we need to check:
But with respect to testing for inclusion of inverses in H (whether for every element $h \in H$, we also have that $h^{-1} \in H)$, we don't need to test for inverses of products of elements, since you've shown $H$ is closed under multiplication.
Indeed, you are correct that for most $h$ with integer entries $a, b$, the inverse $h^{-1} \notin H$ because the corresponding non-zero entries will likely be rational numbers, but not both integers.
So, taking for example $h = \begin{bmatrix} 2 & 0 \\ 0 & 2 \end{bmatrix}$, when we compute $h^{-1} = \begin{bmatrix} \frac 12 & 0 \\ 0 & \frac 12\end{bmatrix}$, we've shown that $h^{-1} \notin H$, because $\frac 12 \notin \mathbb Z$.
Hence, we've shown that $H$ is not closed under taking inverses. (All you need to provide is one counterexample of some element $h \in H$ such that $h^{-1} \notin H$, to show that $H$ is not closed under taking inverses.) And we therefore conclude that $H$, as defined, is NOT a subgroup of $G$. And then you're done.