Welcome to mathimatikoi.org forum; Enjoy your visit here.

True or false statements

Groups, Rings, Domains, Modules, etc, Galois theory
Post Reply
User avatar
Articles: 0
Posts: 170
Joined: Sat Nov 14, 2015 6:32 am
Location: Melbourne, Australia

True or false statements


Post by Riemann » Wed Nov 29, 2017 9:20 am

Let $n \in \mathbb{Z}$ such that $n \geq 2$. Let $\mathcal{S}_n$ be the permutation group on $n$ letters and $\mathcal{A}_n$ be the alternating group. We also denote $\mathbb{C}^*$ the group of non zero complex numbers under multiplication.

Which of the following are correct statements?
  1. For every integer $n \geq 2$ there is a non trivial homomorphism $\chi: \mathcal{S}_n \rightarrow \mathbb{C}^*$.
  2. For every integer $n \geq 2$ there is a unique non trivial homomorphism $\chi:\mathcal{S}_n \rightarrow \mathbb{C}^*$.
  3. For every integer $n \geq 3$ there is a non trivial homomorphism $\chi:\mathcal{A}_n \rightarrow \mathbb{C}^*$.
  4. For every integer $n \geq 5$ there is non trivial homomorphism $\chi:\mathcal{A}_n \rightarrow \mathbb{C}^*$.
Justify your answers.
$\displaystyle \sum_{n=1}^{\infty}\frac{1}{n^s}= \prod_{p \; \text{prime}}\frac{1}{1-p^{-s}}$
Papapetros Vaggelis
Community Team
Community Team
Articles: 0
Posts: 426
Joined: Mon Nov 09, 2015 1:52 pm

Re: True or false statements


Post by Papapetros Vaggelis » Thu Dec 28, 2017 7:09 pm

i. True statement.

If \(\displaystyle{n\in\mathbb{N}\,,n\geq 2}\), then ,we define \(\displaystyle{\mathcal{x}:S_n\to \mathbb{C}^{\star}}\) by

\(\displaystyle{\mathcal{x}(\sigma)=1}\) if \(\displaystyle{\sigma}\) is an even permutation and

\(\displaystyle{\mathcal{x}(\sigma)=-1}\) if \(\displaystyle{\sigma}\) is an odd permutation.

The homomorphism \(\displaystyle{\mathcal{x}}\) is called sign homomorphism.

ii. Without an answer

iv. False statement.

Suppose that iv. is true. Then, if \(\displaystyle{n\in\mathbb{N}\,,n\geq 5}\), we choose a

non trivial homomorphism \(\displaystyle{x_{n}:A_{n}\to \mathbb{C}^{\star}}\), that is \(\displaystyle{\rm{Ker}(x)\neq A_{n}}\).

Since \(\displaystyle{\left(A_{n},\circ\right)}\) is a simple group and \(\displaystyle{\rm{Ker}(x)\trianglelefteq A_{n}}\)

we get \(\displaystyle{\rm{Ker}(x)=\left\{Id\right\}}\), which means that \(\displaystyle{A_{n}\cong Im(x)\leq \mathbb{C}^{\star}}\).

Therefore, \(\displaystyle{A_{n}}\) is cyclic, a contradiction.

So, there exists \(\displaystyle{m\in\mathbb{N}\,,m\geq 5}\) such that, the only homomorphism

\(\displaystyle{x:A_{m}\to \mathbb{C}^{\star}}\) is the trivial one.

iii. False statement

We choose the above \(\displaystyle{m\in\mathbb{N}\,,m\geq 5>3}\).
Post Reply