Problem of the Week #214 - Jul 05, 2016

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Euge

MHB Global Moderator
Staff member
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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Euge

MHB Global Moderator
Staff member
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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