这是一份GLA格拉斯哥大MATHS5077_1作业代写的成功案例

Although the inverse of a homomorphism may not exist, a homomorphism does share some properties in common with isomorphisms. For example, if $\theta: G \rightarrow$ $G^{\prime}$ is a homomorphism, and if $e$ and $e^{\prime}$ are the identities in $G$ and $G^{\prime}$, respectively, then $\theta(e)=\theta\left(e^{2}\right)=\theta(e) \theta(e)$, so that $\theta(e)=e^{\prime}$. Thus a homomorphism maps the identity in $G$ to the identity in $G^{\prime}$. Similarly, as $e^{\prime}=\theta(e)=\theta\left(g g^{-1}\right)=$ $\theta(g) \theta\left(g^{-1}\right)$, we see that for all $g$ in $G$
$$
\theta(g)^{-1}=\theta\left(g^{-1}\right) .
$$
The most obvious example of a homomorphism is a linear map between vector spaces. Indeed, a vector space is a group with respect to addition, and as any linear map $\alpha: V \rightarrow W$ satisfies
$$
\alpha(u+v)=\alpha(u)+\alpha(v)
$$

MATHS5077_1 COURSE NOTES ：

Proof First, every element of $x_{0} K$ is a solution to the given equation, for the general element of $x_{0} K$ is $x_{0} k$, where $k \in K$, and $\theta\left(x_{0} k\right)=\theta\left(x_{0}\right) \theta(k)=y e^{\prime}=$ $y$. Next suppose that $\theta(x)=y$, and consider the element $x_{0}^{-1} x$ in $G$. Then
$$
\theta\left(x_{0}^{-1} x\right)=\theta\left(x_{0}^{-1}\right) \theta(x)=\left(\theta\left(x_{0}\right)\right)^{-1} y=y^{-1} y=e^{\prime},
$$
so that $x_{0}^{-1} x \in K$. This means that $x \in x_{0} K$ and so the set of solutions of $\theta(x)=y$ is the coset $x_{0} K$.