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


*

*$(V,+)$ is an abelian group.

*$1 \in F $ such that $1.\alpha=\alpha$ for every $\alpha \in V$

*$(c_{1}c_{2})\alpha=c_{1}(c_{2}\alpha)$

*$(c_{1}+c_{2})\alpha =c_{1}\alpha +c_{2}\alpha$ 

*$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 $V$ 
 A: 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.
