MATH2601 Higher Linear Algebra (1 Viewer)

InteGrand · Mar 19, 2018

marxman said:
By the same token, how do we prove $a^{2} = b$ in the first place?

$\noindent Well the group is $\{e, a, b\}$ (all \underline{distinct} elements). By closure of the group, $a^2$ must be one of $e, a$, or $b$. Now, what would happen if $a^{2} = e$ or if $a^{2} = a$?$

marxman · Mar 20, 2018

$a^2 = e \implies a^2 b = b \implies a (ab) = b \implies a = b$ . So this cannot be true.
$a^2 = a \implies a = e$ Which also cannot be true, so by elimnation $a^2 = b$

marxman · Mar 20, 2018

Question from 2016 2601 paper:

$\noindent Let $G$ be a group. For any $g \in G$ we define the set \newline \newline $C_g = \{ x \in G \hspace{2mm} | \hspace{2mm} gx = xg \}$ \newline \newline Prove that $C_g$ is a subgroup of $G$. \newline \newline I am not entirely sure where to even begin with this. It seems like a stock standard subgroup question. Normally I would let $x, y \in C_g$ and strive to prove closure and existence of inverse. However with the added condition it seems to complicates things.$

InteGrand · Mar 20, 2018

marxman said:
Question from 2016 2601 paper:

$\noindent Let $G$ be a group. For any $g \in G$ we define the set \newline \newline $C_g = \{ x \in G \hspace{2mm} | \hspace{2mm} gx = xg \}$ \newline \newline Prove that $C_g$ is a subgroup of $G$. \newline \newline I am not entirely sure where to even begin with this. It seems like a stock standard subgroup question. Normally I would let $x, y \in C_g$ and strive to prove closure and existence of inverse. However with the added condition it seems to complicates things.$

$\noindent Perhaps this will make it easier: just redo the question by rewriting the condition in the equivalent form $x^{-1}gx = g$. (And of course, make sure you know / can explain why this is equivalent to the given condition $gx = xg$.) I.e. use $C_{g} = \left\{x\in G : x^{-1}gx = g\right\}$.$

leehuan · Mar 24, 2018

marxman said:
Question from 2016 2601 paper:

$\noindent Let $G$ be a group. For any $g \in G$ we define the set \newline \newline $C_g = \{ x \in G \hspace{2mm} | \hspace{2mm} gx = xg \}$ \newline \newline Prove that $C_g$ is a subgroup of $G$. \newline \newline I am not entirely sure where to even begin with this. It seems like a stock standard subgroup question. Normally I would let $x, y \in C_g$ and strive to prove closure and existence of inverse. However with the added condition it seems to complicates things.$

I'd recommend InteGrand's rearrangement. But it's still fairly easy just using what they give you.

$\text{Closure condition: Let }x,y\in C_g\text{, so we have} \\ \begin{align*} gx &= xg \\ gy &= yg \end{align*}$

$\noindent\text{Then, with the aid of the associative rule,} \\ \begin{align*} g(xy) &= (gx) y\\ &= (xg)y\\ &= x(gy)\\ &= x(yg)\\ &= (xy)g \end{align*}\\ \text{proving that }xy \in C_g\text{ as required.}$

$\noindent\text{Inverses contained condition: For any }x\in C_g\text{ we have }gx = xg.\\ \text{Left and right multiplying by }x^{-1}\text{ gives }x^{-1} g = g x^{-1}\\ \text{which rearranges to }gx^{-1} = x^{-1}g.\\ \text{So }x^{-1}\in C_g\text{ as well.}$

MATH2601 Higher Linear Algebra (1 Viewer)

InteGrand

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marxman

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marxman

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InteGrand

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leehuan

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