1
$\begingroup$

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!

$\endgroup$
  • 1
    $\begingroup$ 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
0
$\begingroup$

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!

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.