Equivalence of presentations of dicyclic group

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This article gives a proof/explanation of the equivalence of multiple definitions for the term dicyclic group
View a complete list of pages giving proofs of equivalence of definitions


Consider the group with presentation:

G := \langle a,b,c \mid a^n = b^2 = c^2 = abc \rangle


  • The element z = a^n = b^2 = c^2 = abc is in the center of G.
  • z^2 is the identity element of G.
  • If we set a = a, x = b^{-1}, and denote by e the identity element, G satisfies the relations a^{2n} = e, x^2 = a^n, xax^{-1} = a^{-1}. Further, the presentation:

\langle a,x \mid a^{2n} = e, x^2 = a^n, xax^{-1} = a^{-1} \rangle

also defines G.

Facts used

  1. Group acts as automorphisms by conjugation


We use the notation as in the statement above.

Step no. Assertion/construction Facts used Given data used Previous steps used Explanation
1 z is central z = a^n = b^2 = c^2 Since every element commutes with its powers, a,b,c all commute with z. Since the group is generated by a,b,c, the centralizer of z is the whole group, so z is central.
2 c = ab, so the group is generated by a,b c^2 = abc We cancel a c from both sides.
3 b = aba b^2 = c^2 Step (2) Combining b^2 = c^2 and Step (2), we get b^2 = abab, so b = aba.
4 b^{-1}ab = a^{-1} Step (3) Multiply both sides of Step (3) by b^{-1} on the left and a^{-1} on the right, then interchange the two sides.
5 b^{-1}a^nb = a^{-n}, i.e., b^{-1}zb = z^{-1} Fact (1) z = a^n Step (4) [SHOW MORE]
6 b^{-1}a^nb = a^n, i.e., b^{-1}zb = z z = a^n = b^2 [SHOW MORE]
7 z = z^{-1} so z^2 is the identity Steps (6),(7)
8 Setting a = a, x = b^{-1}, we get a^{2n} = e, x^2 = a^n, xax^{-1} = a^{-1} z  = a^n = b^2 Steps (4), (7) [SHOW MORE]
9 Conversely, the original relations can be deduced from a^{2n} = e, x^2 = a^n, xax^{-1} = a^{-1} by setting b = x^{-1}, c = ab, so the two presentations define the same group [SHOW MORE]