HOMEWORK ONE: GETTING A FEEL FOR GROUPS
1. Some basics
(a) Show that the empty set does not admit a group structure. (That is, there
does not exist a group G that contains no elements.)
Solution. The identity axiom states that there exists an element e ∈ G such that
ge = eg = g for all g ∈ G. This implies that there exists at least one element in G.
Hence the empty set cannot admit a group structure.
(b) Show that the identity element of a group G is unique. (That is, if two
elements e and e0 satisfy the defining property of the identity element, then
e = e0 .)
Solution. Take two elements e and e0 that satisfy the defining property of the
identity element. In other words, assume that
e0 g = ge0 = g
eg = ge = g,
for all g ∈ G. Then we have ee0 = e0 by taking g = e0 in the first equation, and
ee0 = e by taking g = e in the second equation. This shows that e = ee0 = e0 .
Therefore the identity element of G is unique.
(c) Given an element g ∈ G, show that g −1 is unique. (That is, given elements
h and h0 satisfying the defining property of g −1 , show that h = h0 .)
Solution. Assume that h and h0 satisfy the defining property of g −1 , i.e.,
gh0 = h0 g = e.
gh = hg = e,
Then
h = he = h(gh0 ) = (hg)h0 = eh0 = h0 .
This shows that the inverse g −1 is unique.
(d) Show g = g 2 in a group G if and only if g = e.
Solution. Suppose that g = e. Multiplying g on both sides on the right gives
g 2 = eg = g. Hence g = g 2 .
Conversely, assume that g = g 2 . By the inverse axiom, there exists a g −1 ∈ G
such that g −1 g = e. Multiply g −1 on the left to obtain
e = g −1 g = g −1 g 2 = (g −1 g)g = eg = g.
Thus g = g 2 , and we have proved that g = g 2 if and only if g = e.
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