Permutation does not change if we multiply by left by another group element? Let $(G=(a_1,...,a_n),*)$ be a finite Group. Define for a element $a_i \in G$ a permutation $\phi = \phi(a_i)$ by left multiplication:
$$
 \begin{bmatrix}
a_1 & a_2 & ... & a_n \\
a_i*a_1 & a_i*a_2 & ... & a_i*a_n \\
\end{bmatrix}
$$
I am struggling to understand why this is the permutation as 
$$
 \begin{bmatrix}
a_k*a_1 & a_k*a_2 & ... & a_k*a_n \\
a_i*a_k*a_1 & a_i*a_k*a_2 & ... & a_i*a_k*a_n \\
\end{bmatrix}
$$
where $a_k \in G$. Can somebody give me a reason for why these permutations are the same? Thanks for any help.
 A: We check that the map $\phi: G \times G   \rightarrow G, 
\quad \phi(g_1, g_2) \rightarrow g_1*g_2 $ satisfies the following property:
$$\forall a \in G, \quad h_a(x) = \phi(a, x) = a*x$$ is a bijection on G.  We can see the maps are onto as for any fixed $a \in G$ $$\forall y \in G, \quad \exists a^{-1}y \in G  \text{ such that } h_a(a^{-1}y) = a*(a^{-1}y)=y 
$$
and we can see the maps are injective as for any fixed $a \in G$
$$\begin{align*}
h_a(x) = h_a(y) &\iff \\
a * x = a * y &\iff \\
a^{-1} * (a * x) = a^{-1} * (a * y) &\iff \\
(a^{-1} * a) * x = (a^{-1}*a)  * y  &\iff  \\x = y 
\end{align*}
$$
Thus each map $h_a$ is a bijection on $G$.  Now notice that in your final example the question can be phrased as does the equality $\phi(a_i, x) = \phi(a_i a_k, a_k^{-1} x)$ hold.  Well notice that $$\phi(a_i a_k, a_k^{-1}x) = (a_i a_k)(a_k^{-1} x) = a_i(a_ka_k^{-1})x=a_ix = \phi(a_i, x).$$  Demonstrating that the two permutations are equal.
A: As José has said, both homorphisms have the same effect. Perhaps what you're confused about is that the second homomorphism sends $a_k * a_x \mapsto a_i * a_k * a_x$, seeming like it only applies to elements of the form $a_k * a_x$. While that is true, every element of $G$ may be expressed in the form $a_k * a_x$.
Let $a_x \in G$. Then $a_x = a_k * (a_k^{-1} * a_x)$. The second homomorphism sends
$$a_x = a_k * (a_k^{-1} * a_x)  \mapsto a_i * a_k * (a_k^{-1} * a_x) = a_i * a_x$$
and so is identical to the first homorphism.
