Non-example for vector space

Question

$V$ is a vector space over a field $F$ if it satisfies the following conditions.

1. $(V,+)$ is an abelian group.
2. $1 \in F $ such that $1.\alpha=\alpha$ for every $\alpha \in V$
3. $(c_{1}c_{2})\alpha=c_{1}(c_{2}\alpha)$
4. $(c_{1}+c_{2})\alpha =c_{1}\alpha +c_{2}\alpha$
5. $c(\alpha+\beta)=c\alpha+c \beta $

I want to find examples of sets which are not vector spaces over some Field $F$ by violating exactly one condition among them.

Examples for violating 1 is easy. But what about second one only ? That means $1 \in F$ does not satisfy $1.\alpha=\alpha$ ?

Here scalars $c,c_{1},c_{2}$ are from $F$ and vectors $\alpha$ and $\beta$ are from $V$

Answer 1

Violating only 1:

Be $F = \mathbb F_2$ and $V = \{0,a,b\}$ with the addition table $$\begin{array}{c|ccc} + & 0 & a & b\\ \hline 0 & 0 & a & b\\ a & a & 0 & a\\ b & b & b & 0 \end{array}$$ and the obvious multiplication rule $0\alpha=0$, $1\alpha=\alpha$. Clearly $(V,+)$ is not an abelian group, but one easily verifies that all the other conditions hold.

Violating only 2:

Here I'll take the example proposed by Tobias Kildetoft in the comments:

Be $F$ an arbitrary field, and $(V,+)$ an arbitrary non-trivial abelian group. Decine $c\alpha = 0_V$ for all $c\in F$ and $\alpha \in V$.

Violating only 3:

Take $F=\mathbb C$ and $V=\mathbb R^n$ with the usual per-entry addition (thus fulfilling condition 1). Define $c\alpha = \operatorname{Re}(c)\alpha$ where on the right hand side I've used the usual scalar multiplication with real numbers. Since $\operatorname{Re}(z)$ is linear, conditions 2, 4 and 5 are fulfilled. However if $c_1 = c_2 = \mathrm i$, then the left hand side of condition 3 is $(\mathrm i^2)\alpha = -\alpha$ while the right hand side is $\mathrm i(\mathrm i\alpha) = \mathrm i(\operatorname{Re}(\mathrm i)\alpha) = 0$.

Violating only 4:

Be again $F=\mathbb C$ and $V=\mathbb R^n$ with the usual addition. However now define $c\alpha = \left|c\right|\alpha$.

Violating only 5:

OK, here I'm giving up for now.