Question

# Can the Diffie-Hellman key exchange protocol be extended to three people, Alice, Bob and Carol to...

Can the Diffie-Hellman key exchange protocol be extended to three people, Alice, Bob and Carol to generate session keys? Describe your mathematical argument for why or why not.

Diffie–Hellman key agreement is not limited to a key shared by only two participants. Any number of users can take part in an agreement by performing iterations of the agreement protocol and exchanging intermediate data. For example, Alice, Bob, and Carol could participate in a Diffie–Hellman agreement as follows, with all operations taken to be modulo p:

step 1.The parties agree on the algorithm parameters p and g.
step 2.The parties generate their private keys, named a, b, and c respectively for Alice,Bob and Carol
step 3.Alice computes ga and sends it to Bob.
step 4.Bob computes (ga)b = gab and sends it to Carol.
step 5:Carol computes (gab)c = gabc and uses it as her secret.
step 6:Bob computes gb and sends it to Carol.
step 7:Carol computes (gb)c = gbc and sends it to Alice.
step 8:Alice computes (gbc)a = gabc and uses it as her secret.
step 9:Carol computes gc and sends it to Alice.
step 10:Alice computes (gc)a = gca and sends it to Bob.
step 11:Bob computes (gca)b = gcab = gabc and uses it as his secret.

An eavesdropper has been able to see ga, gb, gc, gab, gac, and gbc, but cannot use any combination of these to efficiently reproduce gabc.

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