Direct product and inverse element of Group

Question

Let $(P_2[\mathbb{R}],+)$ and $(\text{GL}(2,\mathbb R), \cdot)$ be groups. How can I write direct product G of those groups. Let $a = \left (2x^2+4x-3,\begin{pmatrix}3 & 1\\ 2 & 1 \end{pmatrix} \right)$ and $b= \left (-x^2+x-1, \begin{pmatrix} 2 & -1 \\ 1 & 0 \end{pmatrix} \right ).$ Find id element in G, a*b and a^-1.

solution:

So far I have done following:

i) id element in G

$$ \left (0, \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \right ) $$

ii)

$$ a*b= \left (x^2+5x-4, \begin{pmatrix}7 & -3 \\ 5 & -2 \end{pmatrix} \right )$$

Is this the same thing as direct product or I need to do some different steps?

iii)

$$ \left (x^2-x+1, \begin{pmatrix} 1 & -1 \\ -2 & 3 \end{pmatrix} \right) $$

Am I on right way to do this, and is a*b same thing as direct product?

thanks in advance

Answer 1

You have exactly the right idea!

Given two groups $(G,\star)$ and $(H, \cdot)$ we can form their Direct Product $G \times H$ whose elements are pairs $(g,h)$ with $g \in G$ and $h \in H$.

Now this object is a group "componentwise" in the following sense:

• $\text{Id}_{G \times H} = (\text{Id}_G, \text{Id}_H)$
• $(g_1,h_1)(g_2,h_2) = (g_1 \star g_2, h_1 \cdot h_2)$
• $(g,h)^{-1} = (g^{-1}, h^{-1})$

In your case, you correctly identified that, for $G = P_2[\mathbb R]$ and $H = \text{GL}(2,\mathbb{R})$ this means:

• the identity element is $(\text{Id}_G, \text{Id}_H) = \left ( 0 + 0x + 0x^2, \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \right )$
• the product $ab$ is $\left ( (2x^2 + 4x - 3) + (-x^2 + x - 1), \begin{pmatrix} 3 & 1 \\ 2 & 1 \end{pmatrix} \begin{pmatrix} 2 & -1\\1 & 0 \end{pmatrix} \right )$
• the inverse $a^{-1} = \left ( 2x^2 + 4x - 3, \begin{pmatrix} 3 & 1 \\ 2 & 1 \end{pmatrix} \right )^{-1}$ is $\left ( - (2x^2 + 4x - 3), \begin{pmatrix} 3 & 1 \\ 2 & 1 \end{pmatrix}^{-1} \right )$

Notice, in particular, that we write the "inverse" of a polynomial $p$ as $-p$, while we write the inverse of a matrix $A$ as $A^{-1}$. This is because we write the operation on the matrix group as multiplication, but we write polynomial group operation as addition.

As a final remark, you seem to have accidentally taken $b^{-1}$ for part iii. Conceptually, though, your idea is correct.

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I hope this helps ^_^