Cyclic and Dihedral Groups Homework Assignment

Question

My name is David Merlo and I have autism! I am new to asking questions for homework problems for this website, so I am looking forward to learning more of solving math homework problems. I am also new at abstract algebra proofs, so these maybe challenging for me unless I get tutoring on the web. I also have a homework question that I wanted to ask everyone. This Homework question is from Abstract Algebra homework problems which I couldn't find this homework question from the textbook:

Let $G$ be a group with operation $\diamond$ and let $S$ be a set. Suppose $\Phi : G \to S$ is a bijection. Define an operation $⋆ : S × S → S$ by $$ x \star y = \Phi (\Phi^{−1}(x) \diamond \Phi^{−1}(y)) $$ for all $x, y \in S$.

(a.) Show $S$ is a group with respect to $\star$.
(b.) Show that $\Phi : G \to S$ is an isomorphism.

Thank you so much everyone! Have a great day!

Best Regards,

David

Answer 1

Here is a partial solution.

Part a: We wish to show that the set $S$ and its operation $\star:S \times S \to S$ have the properties that define a group structure (AKA the "group axioms"). These properties are

1. associativity, that $(x\star y)\star z = x\star (y\star z)$ holds for all $x,y,z \in S$,
2. the existence of an identity element, i.e. that there exists an element $1_S \in S$ such that $1_S \star x = x \star 1_S = x$ holds for all $x \in S$, and
3. the existence of inverses, i.e. that for every $x \in S$, there exists a corresponding $y \in S$ such that $x \star y = y \star x = 1_S$.

Here is a proof of associativity. Consider any $x,y,z \in S$. We note that $$ \begin{align} (x\star y)\star z &= \Phi(\Phi^{-1}(x)\diamond \Phi^{-1}(y)) \star z \qquad\qquad \qquad \ \ \ \text{(by definition of $\star$)} \\ & = \Phi\left[\Phi^{-1}[\Phi(\Phi^{-1}(x)\diamond \Phi^{-1}(y))] \diamond \Phi^{-1}(z)\right] \quad \text{(by definition of $\star$)} \\ & = \Phi\left[(\Phi^{-1}(x) \diamond \Phi^{-1}(y))\diamond \Phi^{-1}(z)\right] \qquad \quad\ \ \ \text{(by "canceling" $\Phi^{-1}\circ \Phi$)} \\ & = \Phi\left[\Phi^{-1}(x) \diamond (\Phi^{-1}(y)\diamond \Phi^{-1}(z))\right] \qquad \quad \ \ \ \text{(by associativity of $\diamond$)} \\ & = \Phi\left[\Phi^{-1}(x) \diamond \Phi^{-1}[\Phi(\Phi^{-1}(y)\diamond \Phi^{-1}(z))]\right] \\ & = x \star \Phi(\Phi^{-1}(y)\diamond \Phi^{-1}(z)) \\ &= x \star (y \star z). \end{align} $$ Here's a proof of the existence of an identity element. We note that because $G,\diamond$ is a group, $G$ must have an identity element, which is to say that there exists an element $1_G \in G$ such that for every $g \in G$, we have $g\diamond 1_G = 1_G \diamond g = g$. Denote $1_S = \Phi(1_G)$. We can see that $1_S$ must be the identity element of $S$. Indeed, for every $x \in S$, we have \begin{align} 1_S \star x &= \Phi(\Phi^{-1}(x) \diamond \Phi^{-1}(1_S)) \\ & = \Phi(\Phi^{-1}(x) \diamond 1_G) \\ & = \Phi(\Phi^{-1}(x)) \\ & = x. \end{align} Similarly, we can show that $x\star 1_S = x$.

Now, perhaps you can write a similar proof of the existence of inverses.

Part b: In order for $\Phi$ to be an isomorphism, it must be a bijective homomorphism. Since we already know that $\Phi$ is bijective, we only need to show that it is a homomorphism. That is, we must show that for every $g, h \in G$, we have $$ \Phi(g) \star \Phi(h) = \Phi(g \diamond h). $$