What is a non-trivial unit in the integral group ring Z[S_3]?

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    2016
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SUMMARY

A non-trivial unit in the integral group ring $\Bbb Z[S_3]$ can be exemplified by elements that are not simply of the form $1x$ for some $x \in S_3$. The discussion highlights the necessity of understanding the structure of the group ring and the properties of units within it. Specifically, the integral group ring $\Bbb Z[S_3]$ allows for combinations of group elements that yield units not directly tied to individual group elements.

PREREQUISITES
  • Understanding of group theory, specifically symmetric groups like $S_3$
  • Familiarity with integral group rings and their properties
  • Knowledge of units in algebraic structures
  • Basic concepts of ring theory and algebraic structures
NEXT STEPS
  • Study the properties of integral group rings, focusing on $\Bbb Z[S_n]$
  • Explore the concept of units in algebraic structures and their classifications
  • Investigate examples of non-trivial units in other group rings
  • Learn about the applications of group rings in representation theory
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Mathematicians, algebraists, and students studying group theory and ring theory, particularly those interested in the properties of integral group rings and their applications.

Euge
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Here is this week's POTW:

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Give an example of a unit of the integral group ring $\Bbb Z[S_3]$ that is not of the form $1x$ for some $x\in S_3$.

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Remember to read the http://www.mathhelpboards.com/showthread.php?772-Problem-of-the-Week-%28POTW%29-Procedure-and-Guidelines to find out how to http://www.mathhelpboards.com/forms.php?do=form&fid=2!
 
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No one answered this week's problem. You can read my solution below.
Let $a = (12)$ and $b = (123)$. Then $a^2 = b^3 = 1$ and $ab = b^2 a$. Now define

$$g = 1 + a - b^2 + ab - ab^2,$$

$$h = -1 + a + b + ab - ab^2.$$

Then

$$gh$$

$$= -1 + a + b + ab - ab^2 - a + b^2 + ab + b^2 a - a^2 b^2 + b^2 - b^2 a - b^3
- b^2 ab + b^2 ab^2 - ab + aba + ab^2 + ab ab - ab ab^2 + ab^2 - ab^2 a - ab^3 - ab^2 ab + ab^2 ab^2$$

$$= -1 + a + b + ab - ab^2 - a + 1 + ab + b - b^2 + b^2 - ab - 1 - ab^2 + a - ab + b^2 + ab^2 + 1 - b + ab^2 - b - a - b^2 + 1$$

$$= 1$$

and

$$hg$$

$$= -1 - a + b^2 - ab + ab^2 + a + a^2 - ab^2 + a^2 b - a^2 b^2 + b + ba - b^3 + bab - bab^2 + ab + aba - ab^3 + abab - abab^2 - ab^2 - ab^2 a + ab^4 - ab^2 ab + ab^2 ab^2$$

$$= -1 - a + b^2 - ab + ab^2 + a + 1 - ab^2 + b - b^2 + b + ab^2 - 1 + a - ab + ab + b^2 - a + 1 - b - ab^2 - b + ab - b^2 + 1$$

$$= 1.$$

This shows that $g$ and $h$ are multiplicative inverses of each other, neither of which takes the form $1x$ for some $x\in S_3$.
 

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