I'm not sure what definition of supersingular you're taking; I'll assume you mean that the endomorphism ring is an order in a quaternion algebra.

Now, suppose $E$ is a supersingular elliptic curve over $\mathbb{F}_{q}$ of characteristic $p$, and $\varphi$ is the Frobenius endomorphism. You can deduce from noncommutativity of the endomorphism ring (over the algebraic closure) that $[p^n] = \varphi^m$ for some $m,n \in \mathbb{Z}$. The argument goes along the lines of: if $[p^n]$ is never equal to $\varphi^m$, this forces the endomorphism ring to commute as every endomorphism commutes with some power of the Frobenius. You can find more details in Elliptic Curves by Husemoller, including a proof of the converse.

From this, we deduce that multiplication by $p$ is purely inseparable on a supersingular elliptic curve, and hence there is no $p$-torsion.

We can also deduce the following, by degrees of endomorphisms. Let $\alpha,\beta$ be two endomorphisms, then define $\langle \alpha, \beta \rangle = \frac{1}{2} \left ( \deg(\alpha + \beta ) - \deg(\alpha) - \deg(\beta) \right )$, this defines an inner product on the endomorphism ring. Using the above property that $[p^n] = \varphi^m$, some basic algebra shows that $p \mid t$, where $t$ is the trace of the Frobenius, which is equal to $\langle \varphi, 1 \rangle = \\#E(\mathbb{F}_q) - (q+1)$. It is however not necessarily the case that $t=0$; this happens for $q=p$, $p \geqslant 5$ by Hasse's inequality, but there are examples in which $t \neq 0$. Going the other way, we see that $\\#E(\mathbb{F}_q) \equiv 1 \pmod{p}$, so by elementary group theory $E(\mathbb{F}_q)$ has no $p$-torsion, which is equivalent to supersingularity as shown above.

For the question about reduction modulo $p$, the full criterion of Deuring is as follows:

Theorem: Let $L$ be a number field, and $E$ an elliptic curve with complex multiplication by an order in the imaginary quadratic field $K$.
Let $p$ be a (rational) prime, and $P$ a prime above $p$ at which $E$ has good reduction.
Then $E$ has supersingular reduction at $P$ iff there is a unique prime of $K$ above $p$. Otherwise, write $c$ for the conductor of the endomorphism ring of $E$ in $K$, and let $c = c_0 p^k$ with $p \nmid c_0$. Then the ring of endomorphisms of the reduction mod $p$ is $\mathbb{Z} + c_0 \mathcal{O}_K$, the order of $K$ with conductor $c_0$.

I have taken this from Lang's book Elliptic Functions, which contains a proof (page 182 in my edition).

I'm not sure what definition of supersingular you're taking; I'll assume you mean that the endomorphism ring is an order in a quaternion algebra.

Now, suppose $E$ is a supersingular elliptic curve over $\mathbb{F}_{q}$ of characteristic $p$, and $\varphi$ is the Frobenius endomorphism. You can deduce from noncommutativity of the endomorphism ring (over the algebraic closure) that $[p^n] = \varphi^m$ for some $m,n \in \mathbb{Z}$. The argument goes along the lines of: if $[p^n]$ is never equal to $\varphi^m$, this forces the endomorphism ring to commute as every endomorphism commutes with some power of the Frobenius. You can find more details in Elliptic Curves by Husemoller, including a proof of the converse.

From this, we deduce that multiplication by $p$ is purely inseparable on a supersingular elliptic curve, and hence there is no $p$-torsion.

We can also deduce the following, by degrees of endomorphisms. Let $\alpha,\beta$ be two endomorphisms, then define $\langle \alpha, \beta \rangle = \frac{1}{2} \left( \deg(\alpha + \beta ) - \deg(\alpha) - \deg(\beta) \right)$, this defines an inner product on the endomorphism ring. Using the above property that $[p^n] = \varphi^m$, some basic algebra shows that $p \mid t$, where $t$ is the trace of the Frobenius, which is equal to $\langle \varphi, 1 \rangle = \#E(\mathbb{F}_q) - (q+1)$. It is however not necessarily the case that $t=0$; this happens for $q=p$, $p \geq 5$ by Hasse's inequality, but there are examples in which $t \neq 0$. Going the other way, we see that $\#E(\mathbb{F}_q) \equiv 1 \pmod{p}$, so by elementary group theory $E(\mathbb{F}_q)$ has no $p$-torsion, which is equivalent to supersingularity as shown above.

For the question about reduction modulo $p$, the full criterion of Deuring is as follows:

Theorem: Let $L$ be a number field, and $E$ an elliptic curve with complex multiplication by an order in the imaginary quadratic field $K$.
Let $p$ be a (rational) prime, and $P$ a prime above $p$ at which $E$ has good reduction.
Then $E$ has supersingular reduction at $P$ iff there is a unique prime of $K$ above $p$. Otherwise, write $c$ for the conductor of the endomorphism ring of $E$ in $K$, and let $c = c_0 p^k$ with $p \nmid c_0$. Then the ring of endomorphisms of the reduction mod $p$ is $\mathbb{Z} + c_0 \mathcal{O}_K$, the order of $K$ with conductor $c_0$.

I have taken this from Lang's book Elliptic Functions, which contains a proof (page 182 in my edition).

I'm not sure what definition of supersingular you're taking; I'll assume you mean that the endomorphism ring is an order in a quaternion algebra.

Now, suppose $E$ is a supersingular elliptic curve over $\mathbb{F}_{q}$ of characteristic $p$, and $\varphi$ is the Frobenius endomorphism. You can deduce from noncommutativity of the endomorphism ring (over the algebraic closure) that $[p^n] = \varphi^m$ for some $m,n \in \mathbb{Z}$. The argument goes along the lines of: if $[p^n]$ is never equal to $\varphi^m$, this forces the endomorphism ring to commute as every endomorphism commutes with some power of the Frobenius. You can find more details in Elliptic Curves by Husemoller.

From this, we deduce that multiplication by $p$ is purely inseparable on a supersingular elliptic curve, and hence there is no $p$-torsion.

We can also deduce the following, by degrees of endomorphisms. Let $\alpha,\beta$ be two endomorphisms, then define $\langle \alpha, \beta \rangle = \frac{1}{2} \left ( \deg(\alpha + \beta ) - \deg(\alpha) - \deg(\beta) \right )$, this defines an inner product on the endomorphism ring. Using the above property that $[p^n] = \varphi^m$, some basic algebra shows that $p \mid t$, where $t$ is the trace of the Frobenius, which is equal to $\langle \varphi, 1 \rangle = \\#E(\mathbb{F}_q) - (q+1)$. It is however not necessarily the case that $t=0$; this happens for $q=p$, $p \geqslant 5$ by Hasse's inequality, but there are examples in which $t \neq 0$. Going the other way, we see that $\\#E(\mathbb{F}_q) \equiv 1 \pmod{p}$, so by elementary group theory $E(\mathbb{F}_q)$ has no $p$-torsion, which is equivalent to supersingularity as shown above.

For the question about reduction modulo $p$, the full criterion of Deuring is as follows:

Theorem: Let $L$ be a number field, and $E$ an elliptic curve with complex multiplication by an order in the imaginary quadratic field $K$.
Let $p$ be a (rational) prime, and $P$ a prime above $p$ at which $E$ has good reduction.
Then $E$ has supersingular reduction at $P$ iff there is a unique prime of $K$ above $p$. Otherwise, write $c$ for the conductor of the endomorphism ring of $E$ in $K$, and let $c = c_0 p^k$ with $p \nmid c_0$. Then the ring of endomorphisms of the reduction mod $p$ is $\mathbb{Z} + c_0 \mathcal{O}_K$, the order of $K$ with conductor $c_0$.

I have taken this from Lang's book Elliptic Functions, which contains a proof (page 182 in my edition).

I'm not sure what definition of supersingular you're taking; I'll assume you mean that the endomorphism ring is an order in a quaternion algebra.

Now, suppose $E$ is a supersingular elliptic curve over $\mathbb{F}_{q}$ of characteristic $p$, and $\varphi$ is the Frobenius endomorphism. You can deduce from noncommutativity of the endomorphism ring (over the algebraic closure) that $[p^n] = \varphi^m$ for some $m,n \in \mathbb{Z}$. The argument goes along the lines of: if $[p^n]$ is never equal to $\varphi^m$, this forces the endomorphism ring to commute as every endomorphism commutes with some power of the Frobenius. You can find more details in Elliptic Curves by Husemoller, including a proof of the converse.

From this, we deduce that multiplication by $p$ is purely inseparable on a supersingular elliptic curve, and hence there is no $p$-torsion.

We can also deduce the following, by degrees of endomorphisms. Let $\alpha,\beta$ be two endomorphisms, then define $\langle \alpha, \beta \rangle = \frac{1}{2} \left ( \deg(\alpha + \beta ) - \deg(\alpha) - \deg(\beta) \right )$, this defines an inner product on the endomorphism ring. Using the above property that $[p^n] = \varphi^m$, some basic algebra shows that $p \mid t$, where $t$ is the trace of the Frobenius, which is equal to $\langle \varphi, 1 \rangle = \\#E(\mathbb{F}_q) - (q+1)$. It is however not necessarily the case that $t=0$; this happens for $q=p$, $p \geqslant 5$ by Hasse's inequality, but there are examples in which $t \neq 0$. Going the other way, we see that $\\#E(\mathbb{F}_q) \equiv 1 \pmod{p}$, so by elementary group theory $E(\mathbb{F}_q)$ has no $p$-torsion, which is equivalent to supersingularity as shown above.

For the question about reduction modulo $p$, the full criterion of Deuring is as follows:

Theorem: Let $L$ be a number field, and $E$ an elliptic curve with complex multiplication by an order in the imaginary quadratic field $K$.
Let $p$ be a (rational) prime, and $P$ a prime above $p$ at which $E$ has good reduction.
Then $E$ has supersingular reduction at $P$ iff there is a unique prime of $K$ above $p$. Otherwise, write $c$ for the conductor of the endomorphism ring of $E$ in $K$, and let $c = c_0 p^k$ with $p \nmid c_0$. Then the ring of endomorphisms of the reduction mod $p$ is $\mathbb{Z} + c_0 \mathcal{O}_K$, the order of $K$ with conductor $c_0$.

I have taken this from Lang's book Elliptic Functions, which contains a proof (page 182 in my edition).