[Math] To which group is $G/\ker \phi$ isomorphic to

abstract-algebragroup-theory

$G = \langle H, \odot_7\rangle$ where $H= \{ 1,2,3,4,5,6\}$ and $\odot_7$ denotes the operation, multiplication modulo $7$, the function $\phi: G \to G$ defined by $\phi(g)=g^2$ .
List the elements of $G/\ker(\phi)$. To which group is $G/\ker(\phi)$ isomorphic to?

I am confused on what to do from here. I know that $G= \{1,2,3,4,5,6\}$ under multiplication modulo $7$, and $ker(\phi)= \{1,6\}$. so we would have $G/\{1,6\}$. Apparently the answer is:
$\{\{1,6\},\{2,5\},\{3,4\}\}$. Can someone please explain to me why this is and also the general process I would take to get this given $ker(\phi)$ and $G$? Also, from here how would I find what this is isomorphic to? I know that it is isomorphic to $\phi(G)$ but I don't know why…

Best Answer

As you've shown $Ker(\phi)=\{1,6\}$

You'd have $G/Ker(\phi)=\{\{1,6\},\{a,b\},\{c,d\}\}$

Note that $Ker(\phi)$ partitions the set $G$ in such a way that the corresponding pairs which are formed map to the same element under $\phi$.

Why?

By definition of cosets $G/Ker(\phi)=\{g.Ker(\phi)\,\,\big| \,\,g\in G\}$, then $\{a,b\}=\{g.1,g.6\}$ for some $g$.

Now $\phi(a)=\phi(g.1)=\phi(g)\phi(1)=\phi(g)\phi(6)=\phi(g.6)=\phi(b)$

Recall: $\phi(1)=1=\phi(6)$. Also note that $\phi(2)=4=\phi(5)$ and $\phi(3)=2=\phi(4)$ (NOTE that the operations still take place modulo $7$).

How did we reach here?

Well it is a well known fact that when we talk about ordinary addition $g^2=(-g)^2$

Now, same is the case here (with a little twist), what is the additive (modulo 7) inverse of $2$? $5$ is it? Of course, in fact additive inverse of $a$ (modulo 7) is $7-a$ and you have the same concept as the one used above.

$$a^2\equiv (7-a)^2\pmod7$$

So $\{a,7-a\}\in G/Ker(\phi)$ for $a\in H$ and the set becomes

$G/Ker(\phi)=\{\{1,6\},\{2,5\},\{3,4\}\}$

There is a more rigorous way that has been shown in most of the answers above, this is slightly different (intuitive?) way

Now, for the part where you ask what is $G/Ker(\phi)$ isomorphic to

Note $|G/Ker(\phi)|=|G|/|Ker(\phi)|=3$ and as it is a well known fact that any group (of quotients here) of prime order is cylic. So, $G/Ker(\phi)$ is isomorphic to any cyclic group of order 3. There is a theorem (Fundamental Theorem of Isomorphism) which states that

If $\phi:G\to G'$ (here G=G') is a homomorphism (Note that $G$ is onto $\phi(G)$), then $G/Ker(\phi)\cong\phi(G)$.

Now note, here $\phi(G)=\{1,2,4\}=\langle 4\rangle=\langle 2\rangle$ (modulo 7)

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