Answer to Question #136600 in Abstract Algebra for emmanuel ehi

Question

(a) Let M and T be a groups and φ : M −→ T an epimorphism. Prove that if M is Abelian then T is Abelian.
(b) Let G be a group and let H be a subgroup of index 2. Prove that for every g ∈ G, g−1 / ∈ H whenever g / ∈ H. Hence, prove that for all a,b ∈ G, if a / ∈ H and b / ∈ H then ab ∈ H.
(c) Find all the distinct left cosets of the subgroup H = h(1,1)i in Z2 ×Z4.

Answer 1

(a) Let $a,b\in T$ be arbitrary. Since $\varphi : M\to T$ is epimorphism, there are exist $x,y\in M$ such that $\varphi(x)=a, \varphi(y)=b.$ Taking into account that $xy=yx$ for any $x,y\in M$ we conclude that

$ab=\varphi(x)\varphi(y)=\varphi(xy)=\varphi(yx)=\varphi(y)\varphi(x)=ba.$

Therefore, T is Abelian.

(b) If $g\notin H,$ then $g^{-1}\notin H.$ Indeed, suppose that $g^{-1}\in H$. Since $H$ is a group, $g=(g^{-1})^{-1}\in H$ and we have a contradiction.

Each subgroup $H$ of index 2 is normal in $G$ because $g_1H=G\setminus H=Hg_2$ for any $g_1\notin H, g_2\notin H.$ Consequenly, the quotient group $G/H$ exists and is of order 2. Thus, $(gH)(gH)=H$ for any $g\notin H.$ If $a\notin H$ and $b\notin H$, then $a,b\in G\setminus H=gH$ for any $g\notin H$, and therefore, $ab\in(gH)(gH)=H.$

(с) Let $H=\langle(1,1)\rangle$. Taking into account that $(1,1)+(1,1)=(0,2), (1,1)+(1,1)+(1,1)=(1,3)$ and $(1,1)+(1,1)+(1,1)+(1,1)=(0,0)$ we conclude that

$H=\{(1,1), (0,2),(1,3),(0,0)\}$.

Let us find all the distinct left cosets:

$[(0,0)]=(0,0)+H=H,$

$[(0,1)]=(0,1)+H=\{(1,2), (0,3),(1,0),(0,1)\}.$

Since $[(0,0)]\cup[(0,1)]=\mathbb Z_2\times\mathbb Z_4,$ there are 2 distinct left coset of the subgroup $H$ in $\mathbb Z_2\times\mathbb Z_4$: $[(0,0)]$ and $[(0,1)].$