Yes, it is true. Clearly, it suffices to consider the case that the representation is faithul. Let $G$ be the cyclic group, and $h \in G $ be an element of order $p$. By Clifford's Theorem and Machke's Theorem ( and since $G$ is Abelian), the representation restricts to $\langle h \rangle$ as a direct sum of equivalent faithful irreducible representations. It suffices to prove that each of these has dimension $p-1$. Hence it suffices to prove that any faithful irreducible $\mathbb{Q}_{p}$-representation of $\langle h \rangle $ has dimension $p-1$.

Let $V$ be the underlying module. Then the minimum polynomial of $h$ on $V$ divides $x^{p}-1$, but does not have $x-1$ as a factor, so the minimum polynomial of $x$ on $V$ divides $1 + x + \ldots + x^{p-1}$. But the proof that this polynomial is irreducible over $\mathbb{Q}$ using Eisenstein's criterion applies equally well to show that the polynomial is irreducible over $\mathbb{Q}_{p}$.

Now $V$ is spanned by $\{vh^{i} : 0 \leq i \leq p-2\}$ for any non-zero $v \in V$ by irreducibility, so ${\rm dim}(V) \leq p-1$. But we now also know that $\frac{x^{p}-1}{x-1}$ is the minimum polynomial of $x$ on $V$, so certainly ${\rm dim} (V) \geq p-1$. Hence ${\rm dim} (V) = p-1$, as required.

Yes, it is true. Clearly, it suffices to consider the case that the representation is faithul. Let $G$ be the cyclic group, and $h \in G $ be an element of order $p$. By Clifford's Theorem and Maschke's Theorem ( and since $G$ is Abelian), the representation restricts to $\langle h \rangle$ as a direct sum of equivalent faithful irreducible representations. It suffices to prove that each of these has dimension $p-1$. Hence it suffices to prove that any faithful irreducible $\mathbb{Q}_{p}$-representation of $\langle h \rangle $ has dimension $p-1$.

Let $V$ be the underlying module. Then the minimum polynomial of $h$ on $V$ divides $x^{p}-1$, but does not have $x-1$ as a factor, so the minimum polynomial of $x$ on $V$ divides $1 + x + \ldots + x^{p-1}$. But the proof that this polynomial is irreducible over $\mathbb{Q}$ using Eisenstein's criterion applies equally well to show that the polynomial is irreducible over $\mathbb{Q}_{p}$.

Now $V$ is spanned by $\{vh^{i} : 0 \leq i \leq p-2\}$ for any non-zero $v \in V$ by irreducibility, so ${\rm dim}(V) \leq p-1$. But we now also know that $\frac{x^{p}-1}{x-1}$ is the minimum polynomial of $x$ on $V$, so certainly ${\rm dim} (V) \geq p-1$. Hence ${\rm dim} (V) = p-1$, as required.

Yes, it is true. Clearly, it suffices to consider the case that the representation is faithul. Let $G$ by the cyclic group, and $h \in G $ be an element of order $p$. By Clifford's Theorem and Machke's Theorem ( and since $G$ is Abelian), the representation restricts to $\langle h \rangle$ as a direct sum of equivalent faithful irreducible representations. It suffices to prove that each of these has dimension $p-1$. Hence it suffices to prove that any faithful irreducible $\mathbb{Q}_{p}$-representation of $\langle h \rangle $ has dimension $p-1$.

Let $V$ be the underlying module. Then the minimum polynomial of $h$ on $V$ divides $x^{p}-1$, but does not have $x-1$ as a factor, so the minimum polynomial of $x$ on $V$ divides $1 + x + \ldots + x^{p-1}$. But the proof that this polynomial is irreducible over $\mathbb{Q}$ using Eisenstein's criterion applies equally well to show that the polynomial is irreducible over $\mathbb{Q}_{p}$.

Now $V$ is spanned by $\{vh^{i} : 0 \leq i \leq p-1 \}$ for any non-zero $v \in V$ by irreducibility, so ${\rm dim}(V) \leq p-1$. But we now also know that $\frac{x^{p}-1}{x-1}$ is the minimum polynomial of $x$ on $V$, so certainly ${\rm dim} (V) \geq p-1$. Hence ${\rm dim} (V) = p-1$, as required.

Yes, it is true. Clearly, it suffices to consider the case that the representation is faithul. Let $G$ be the cyclic group, and $h \in G $ be an element of order $p$. By Clifford's Theorem and Machke's Theorem ( and since $G$ is Abelian), the representation restricts to $\langle h \rangle$ as a direct sum of equivalent faithful irreducible representations. It suffices to prove that each of these has dimension $p-1$. Hence it suffices to prove that any faithful irreducible $\mathbb{Q}_{p}$-representation of $\langle h \rangle $ has dimension $p-1$.

Let $V$ be the underlying module. Then the minimum polynomial of $h$ on $V$ divides $x^{p}-1$, but does not have $x-1$ as a factor, so the minimum polynomial of $x$ on $V$ divides $1 + x + \ldots + x^{p-1}$. But the proof that this polynomial is irreducible over $\mathbb{Q}$ using Eisenstein's criterion applies equally well to show that the polynomial is irreducible over $\mathbb{Q}_{p}$.

Now $V$ is spanned by $\{vh^{i} : 0 \leq i \leq p-2\}$ for any non-zero $v \in V$ by irreducibility, so ${\rm dim}(V) \leq p-1$. But we now also know that $\frac{x^{p}-1}{x-1}$ is the minimum polynomial of $x$ on $V$, so certainly ${\rm dim} (V) \geq p-1$. Hence ${\rm dim} (V) = p-1$, as required.