In order to construct an example of Herbrand quotient, I want to know the cohomology group of $H^i(\mathbb{Z}/p\mathbb{Z}, \mathbb{Z})$ and $H^i(\mathbb{Z}/p\mathbb{Z}, \mathbb{Z}/p\mathbb{Z})$ for $i = 0, 1$.
When $i = 0$, I know $H^0(G, M) = M^G$.
Therefore, I guess $H^0(\mathbb{Z}/p\mathbb{Z}, \mathbb{Z}) = \{x \in \mathbb{Z} \mid \sigma + x = x, x \in \mathbb{Z}/p\mathbb{Z}\} = p\mathbb{Z}$, and $H^0(\mathbb{Z}/p\mathbb{Z}, \mathbb{Z}/p\mathbb{Z}) = \{x \in \mathbb{Z}/p\mathbb{Z} \mid \sigma + x = x, x \in \mathbb{Z}/p\mathbb{Z}\} = p\mathbb{Z}$. Is this right?
But I cannot calculate the case of $i = 1$.
I would appreciate if you could help me calculating $H^1(\mathbb{Z}/p\mathbb{Z}, \mathbb{Z}/p\mathbb{Z})$ and $H^1(\mathbb{Z}/p\mathbb{Z}, \mathbb{Z})$. Thank you.
You need to specify the action of $C_p=\Bbb Z/p\Bbb Z$ on $M$ in order to make $H^i(C_p,M)$ meaningful. That is you need an automorphism $\alpha$ of $M$ with $\alpha^p$ equalling the identity. Then, for a generator $g$ of $C_p$ you define $m\cdot g=\alpha(m)$.
In any case, one can take any Abelian group $M$ and the trivial action of $C_p$, that is $m\cdot g=m$. Even in this case, the cohomology is interesting. In this case $H^0(C_p,M)=M^G=M$. The Tate cohomology (with occurs in the Herbrand quotient) is slightly more interesting: $\hat H^0(C_p,M)=M^G/T(M)$ where $T$ is the trace map: $$T(m)=\sum_{k=0}^{p-1}m\cdot g^k.$$ When the action is trivial, then $T(m)=pm$, and so $\hat H^0(C_p,M)=M/pM$. In particular, $\hat H^0(C_p,\Bbb Z)\cong \Bbb Z/p\Bbb Z$ and $\hat H^0(C_p,\Bbb Z/p\Bbb Z)\cong \Bbb Z/p\Bbb Z$.
What about $H^1$. For the cyclic group $C_p$, $$H_1(C_p,M)\cong\frac{\ker T}{\{m-m\cdot g:m\in M\}}.$$ When the action on $M$ is trivial, the denominator vanishes and $T$ is multiplication by $p$ so that $$H_1(C_p,M)\cong\{m\in M:pm=0\}.$$ In particular, $H^1(C_p,\Bbb Z)=\{0\}$ and $H^1(C_p,\Bbb Z/p\Bbb Z)\cong \Bbb Z/p\Bbb Z$.