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$G = \langle H, \odot_7\rangle$ 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...

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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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To get the kernel, you need to find all those elements $g\in G$ such that $\phi(g)=1$ (since $1$ is the identity of the group $G$. In your case, this corresponds to finding $g$ such that $g^2 \equiv 1\pmod{7}$. This yields the solutions for kernel $K=\{1,6\}$.

Now $G/K$ is the set of (left or right) cosets of $K$ in $G$. Since everything is Abelian hence normal, thus it forms a quotient group. Since the order of $G$ Is 6 and that of $K$ is 2, by Lagrange's theorem the order of $G/K$ will be 3.

The cosets of $K$ are : $K, 2K, 3K$. You should write down the elements of these sets to see what you obtain.

Now that we have a group of order $3$ (prime), hence cyclic, hence isomorphic to $\mathbb{Z/3Z}$.

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This is just the most simple application of the first isomorphism theorem for groups, which, as you surely know, states that if $\phi:G\longrightarrow G$ is a homomorphism, then $G/Ker(\phi)\cong \phi(G)$. To list explicitely the elements of $G/Ker(\phi)$ it is enough to compute $a*Ker(\phi)$ for each $a\in G$. In your example, if you do $a*Ker(\phi)$ for $a=1,2,\cdots,6$ you'll get:

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

Then $G/Ker(\phi)=\{a*Ker(\phi):a\in G\}=\{\{1,6\}, \{2,5\}, \{3,4\}\}$

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So you have $G/\{1, 6\}$.

The elements of a quotient group $G/H$ are the cosets of the group $H$. The cosets of $H$ are found by taking some element $g \in G$ and computing $gH$. If you try this with every element of $G$, you will obtain 3 cosets. Let $K = \{1, 6\}$

$1K = \{1, 6\}$

$2K = \{2, 5\}$

$3K = \{3, 4\}$

$4K = \{4, 3\} = 3K$

$5K = \{5, 2\} = 2K$

$6K = \{6, 1\} = 1K$

So the unique cosets of $K$ are $\{K, 2K, 3K\}$, and these are the elements of $G/K$.

Just by knowing that $G/K$ is of order 3, you can conclude that it is isomorphic to $\mathbb{Z}_3$ - the cyclic group of order 3.

Also, the first isomorphism theorem says:

$G/ker(\varphi) \cong Im(\varphi)$

And the image of your mapping $\varphi$ is $\{1, 2, 4\}$, a group of order 3.

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