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Quaternary group ring codes: Ranks, kernels and self-dual codes

This work has been partially supported by the Spanish MINECO under Grant TIN2016-77918-P (AEI/FEDER, UE)

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  • We study $ G $-codes over the ring $ {\mathbb{Z}}_4 $, which are codes that are held invariant by the action of an arbitrary group $ G $. We view these codes as ideals in a group ring and we study the rank and kernel of these codes. We use the rank and kernel to study the image of these codes under the Gray map. We study the specific case when the group is the dihedral group and the dicyclic group. Finally, we study quaternary self-dual dihedral and dicyclic codes, tabulating the many good self-dual quaternary codes obtained via these constructions, including the octacode.

    Mathematics Subject Classification: Primary: 16S34; Secondary: 94B15.

    Citation:

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  • Table 1.  (Extremal) Dihedral Self-dual Codes of length 4

    First Row of A First row of B Min Lee Weight Lee Weight Distribution
    (1, 1) (1, 3) 4 $ 1+14z^4+z^{8} $
    (1, 1) (3, 1) 4 $ 1+14z^4+z^{8} $
    (1, 3) (1, 1) 4 $ 1+14z^4+z^{8} $
    (1, 3) (3, 3) 4 $ 1+14z^4+z^{8} $
    (3, 1) (1, 1) 4 $ 1+14z^4+z^{8} $
    (3, 1) (3, 3) 4 $ 1+14z^4+z^{8} $
    (1, 3) (1, 3) 4 $ 1+14z^4+z^{8} $
    (3, 3) (3, 1) 4 $ 1+14z^4+z^{8} $
     | Show Table
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    Table 2.  (Extremal) Dihedral Self-dual Codes of length 8

    First Row of $ A $ First row of $ B $ Min Lee Weight Lee Weight Distribution
    (0, 0, 2, 2) (1, 1, 3, 1) 4 $ 1+28z^4+198z^{8}+\dots $
    (0, 0, 0, 0) (1, 3, 1, 1) 4 $ 1+28z^4+198z^{8}+\dots $
    (0, 0, 0, 2) (3, 1, 3, 1) 4 $ 1+12z^4+64z^{6}+102z^8+\dots $
    (0, 0, 2, 0) (1, 1, 3, 3) 4 $ 1+12z^4+64z^{6}+102z^8+\dots $
     | Show Table
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    Table 3.  Best Dicyclic Self-dual Codes of lengths 4, 8 and 12

    $ n $ 1st row of $ A $ 1st row of $ B $ 1st row of $ C $ Min Lee Weight Gray Image Linear
    $ 4 $ (1, 3) (3, 3) (3, 3) 4 Yes
    $ 8 $ (0, 0, 0, 2) (3, 3, 3, 3) (3, 3, 3, 3) 4 Yes
    $ 8 $ (0, 0, 1, 1) (0, 0, 1, 3) (1, 3, 0, 0) 4 No
    $ 8 $ (0, 0, 1, 1) (0, 1, 1, 2) (1, 2, 0, 1) $ 6^* $ No
    $ 12 $ (0, 0, 0, 0, 0, 0) (0, 1, 3, 0, 1, 1) (0, 1, 1, 0, 1, 3) 4 Yes
     | Show Table
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