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The group of units of a field acts on the field by left multiplication

Solution to Abstract Algebra by Dummit & Foote 3rd edition Chapter 1.7 Exercise 1.7.1

Solution: Let $a \in F$. We have $1 \cdot a = 1a = a$ by the definition of identity. Now let $g_1, g_2 \in F^\times$. Then $$g_1 \cdot (g_2 \cdot a) = g_1 \cdot g_2 a = g_1(g_2a) = (g_1g_2)a = (g_1g_2) \cdot a,$$ by the associativity of multiplication in $F$. So left multiplication by nonzero elements is in fact a group action.


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This Post Has 2 Comments

  1. Is it necessary to additionally show for the case that "a" equals the identity of additive group of the field (i.e. a = 0)? since if a = 0, then "a" is not in the multiplicative group of the field and we thus cannot use the group axioms the multiplicative group such as associativity and identity.

    1. No. There is no need to consider $a=0$ as $a\notin F^\times$.

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