# MATH2601 Higher Linear Algebra (1 Viewer)

#### InteGrand

##### Well-Known Member
By the same token, how do we prove $\bg_white a^{2} = b$ in the first place?
$\bg_white \noindent Well the group is \{e, a, b\} (all \underline{distinct} elements). By closure of the group, a^2 must be one of e, a, or b. Now, what would happen if a^{2} = e or if a^{2} = a?$

#### marxman

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

#### marxman

##### Member
Question from 2016 2601 paper:

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

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#### InteGrand

##### Well-Known Member
Question from 2016 2601 paper:

$\bg_white \noindent Let G be a group. For any g \in G we define the set \newline \newline C_g = \{ x \in G \hspace{2mm} | \hspace{2mm} gx = xg \} \newline \newline Prove that C_g is a subgroup of G. \newline \newline I am not entirely sure where to even begin with this. It seems like a stock standard subgroup question. Normally I would let x, y \in C_g and strive to prove closure and existence of inverse. However with the added condition it seems to complicates things.$
$\bg_white \noindent Perhaps this will make it easier: just redo the question by rewriting the condition in the equivalent form x^{-1}gx = g. (And of course, make sure you know / can explain why this is equivalent to the given condition gx = xg.) I.e. use C_{g} = \left\{x\in G : x^{-1}gx = g\right\}.$

#### leehuan

##### Well-Known Member
Question from 2016 2601 paper:

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

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

\bg_white \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.}

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