Kernel: SageMath 9.8
The group (the binary tetrahedral group) is not the group of units in a ring of characteristic 4.
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x = 1+c+i
x^24 = 1
x^2 = 1+2*c+2*i+-1+c^2+ci+cj
x^3 = 2*1-c-i--1-c^2-ci-cj+2*-c+-i+c^2i+c^2j+c^2k+-ck
(i+j+k-1)^2 = 1+i+j--1+k--i--j--k
(i+j+k)^2 = i+j--1+k+-i+-j+-k
Suppose has characteristic 4 and realizes In what follows, we make use of the computations in the above cell which occur in
Working mod , we know that since is the unique element of order 2 in . Since we have that For any , , so either or mod . But since is a unit and in we must have mod for all . Using all this, and the above computation of we see that If we obtain so and has order 1 or 3. If then again has order 1 or 3. We check all cases below and we see that is never realized. (The below computations show that the natrual map cannot be injective, a contradiction.)
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i + j + k + 1 cases:
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x = 1
S has order 2 and characteristic 2
S* = 1
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x = c
S has order 8 and characteristic 2
S* = C3
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x = c^2
S has order 2 and characteristic 2
S* = 1
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x = ci
S has order 32 and characteristic 2
S* = A4
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x = cj
S has order 32 and characteristic 2
S* = A4
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x = ck
S has order 32 and characteristic 2
S* = A4
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x = -c^2i
S has order 2 and characteristic 2
S* = 1
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x = -c^2j
S has order 2 and characteristic 2
S* = 1
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x = -c^2k
S has order 2 and characteristic 2
S* = 1
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i + j + k - 1 cases:
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x = 1
S has order 2 and characteristic 2
S^* = 1
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x = c
S has order 8 and characteristic 2
S^* = C3
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x = c^2
S has order 2 and characteristic 2
S^* = 1
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x = ci
S has order 32 and characteristic 2
S^* = A4
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x = cj
S has order 32 and characteristic 2
S^* = A4
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x = ck
S has order 32 and characteristic 2
S^* = A4
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x = -c^2i
S has order 2 and characteristic 2
S^* = 1
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x = -c^2j
S has order 2 and characteristic 2
S^* = 1
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x = -c^2k
S has order 2 and characteristic 2
S^* = 1
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