I am trying to find a group that is isomorphic to the quotient group G/H given G = $\mathbb({Z7}$-{0},$\times7)$ where the normal subgroup G = {1,6}.

I have found that the quotient group G/H is composed of {{1,6},{2,5},{3,4}} with order 3 but I am now struggling with the next part. I know that for a group to be isomorphic to G/H it must be bijective and be a homomorphism but I am not sure how to formulate one by myself, thanks!

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    $\begingroup$ What's $(Z7,\times 7)$? $\endgroup$ – user228113 Nov 19 '17 at 15:10
  • $\begingroup$ hint: know of any good isomorphism theorems? $\endgroup$ – thesmallprint Nov 19 '17 at 15:11
  • $\begingroup$ @G.Sassatelli Sorry I couldn't find the appropriate MathJax formating, it's meant to be the set 1-6 with the binary operation multiplication mod 7, thanks! $\endgroup$ – Zombiegit123 Nov 19 '17 at 15:15
  • $\begingroup$ @thesmallprint I know the first 3 but am not sure which one to apply, sorry. $\endgroup$ – Zombiegit123 Nov 19 '17 at 15:16
  • $\begingroup$ I would write $(\mathbb Z/7\mathbb Z)^\times$ $\endgroup$ – neptun Nov 19 '17 at 15:19

There is really "only one" group with three elements, $\mathbb Z/3\mathbb Z$. You just need to find out which element in your group is the neutral element and how the other two behave when multiplied, and you can easily write down a 1-1 correspondence, which if you verify that it is a group homomorphism will be your isomorphism.

Edit. Some very explicit details.

Let the three elements of the group $G/H$ be denoted $a,b,c$.

1. Investigating the structure.

Note that mod 7 we have $$\text{a}^2:\quad 1^2 = 6^2= 1$$ $$\text{b}^2:\quad3^2 = 4^2 = 2$$ $$\text{c}^2:\quad2^2 = 5^2 = 4$$ with the additional properties that $1*1=1$ and $2*2=4$ and $2*4= 1$.

Let's write these as $a*a=a$ and $b*b=c$ and $b*c=a$.

This is just like $\mathbb Z/3$ (with addition) where $0+0=0$ and $1+1=2$ and $1+2=0$.

2. Defining a bijection.

Let $f:G/H\to \mathbb Z/3$ be defined by $1\mapsto 0$, $3\mapsto 1$ and $2\mapsto 4$. It is obviously bijective. You just need to verify that it is a homomorphism!

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  • $\begingroup$ Can you provide any more hints? I'm stumped. $\endgroup$ – Zombiegit123 Nov 19 '17 at 16:04
  • $\begingroup$ @Zombiegit123 added some details! $\endgroup$ – neptun Nov 19 '17 at 21:47

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