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I know for a fact that DES is not a group, but are any of the Triple-Des versions a group? Why, or why not?

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    $\begingroup$ Keith Campbell & Michael Wiener's paper proving that DES is not a group is (currently) available for free from the publisher, in a much better format than I once paid for. $\endgroup$
    – fgrieu
    Jun 28, 2013 at 11:14

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No, they aren't a group.

Justification: We know that the subgroup generated by DES is very large. If (any of the variants of) 3DES formed a closed group, then the subgroup generated by 3DES would be no larger than $2^{168}$. We know the latter is not the case. Therefore, the former is not the case, either.

Also, for the EDE variants of 3DES, it is easy to see that the size of the subgroup generated by 3DES is the same as the size of the subgroup generated by DES (consider what happens with EDE-3DES when you set all three subkeys equal to each other; the result is equivalent to plain DES).

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    $\begingroup$ I initialy wondered if "the subgroup generated by DES" is the subgroup of the group of (even) permutations generated by the set of (likely, $2^{56}$) permutations defined by DES encryption; or by the set of (likely, $2^{57}-4$) permutations defined by DES encryption and decryption. Now I realize these definitions are equivalent. $\endgroup$
    – fgrieu
    Jun 28, 2013 at 9:11

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