[Math] Groups under Multiplication

abstract-algebragroup-theorymatrices

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:

  • $(1)$ the identity element of $G$, namely $e = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$ is in $H$ and that
  • $(2)$ for two elements $h_1, h_2 \in H$, $h_1\cdot h_2 \in H$,

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.

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