Is DDH hard over this group? I'm new to DDH. 
Reading this survey, I noticed that DDH is (believed to be) hard in many groups, but most of them are prime-order groups (the only one that is not is the cyclic subgroup of order $(p-1)(q-1)$ of the group of integers modulo $N = pq$).
My question is about the hardness of DDH in this specific composite-order group:


  Let $q$ be a prime such that $p = 4q+1$ is a prime. Is DDH hard in the subgroup of order $2q$ of $\Bbb Z_{p}^\ast$?




Note about this group: My algebra background is fairly basic and I don't even know if this group is guaranteed to exist, and if so, if it guaranteed to be cyclic.
Is DDH hard in the subgroup of order $2q$ of $\mathbb{Z}_p^*$


No, it is not.

If you define the function $F(x) = x^{q} \bmod p$, then for any $x$ is\n the group of order $2q$, we have $F(x)= 1$ if $x$ is a member of the proper subgroup of order $q$, and $-1$ if it is not.  And, exactly half of the elements will be in that subgroup.

You will always have an even number (either 0 or 2) of $F(g^a)$, $F(g^b)$, $F(g^{ab})$ be -1; if you find that an odd number of $F(g^a)$, $F(g^b)$, $F(g^c)$ be -1, then you have determined that is not a DH triple.

This logic applies to any group that has an order with a small factor.