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I know this is a basic question, but I'm trying to convince myself of Wikipedia's statement. "The quotient group $G / G$ is isomorphic to the trivial group."

I write the definition for left multiplication because left cosets = right cosets. $ G/G = \{g \in G : gG\} $ But how is this isomorphic to the trivial group, $ \{id_G\} $? $gG$ can't be simplified to $id_G$ ?

Thank you.

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10  
How many cosets are there? –  Chris Eagle Dec 28 '12 at 21:04
5  
Isomorphic, not equal. –  Hurkyl Dec 28 '12 at 21:11

7 Answers 7

up vote 5 down vote accepted

Well, forgive me for getting all technical, but this isn't right: $G/G = \{ g \in G \colon gG\}$. It should be the other way around: $G/G = \{ gG |\, g \in G \}$.

Next, we get rid of $g$ by realizing that $gG = G$ for any $g \in G$. Therefore, $G/G = \{G\}$, i.e. $G/G$ is a set with exactly 1 element, and this element is $G$ itself.

Now, the next thing is to realize how the operation in $G/G = \{G\}$ works. It works like this: $G \cdot G = G$. Exactly like that of the trivial group $\{1\}$: $1 \cdot 1 = 1$. So, the map $G/G \to \{1\}$ that sends $G$ to $1$ is a group isomorphism.

I know, I know, this may be the worst way to explain these things. There are way too many trivial formulas there. But it can be good for convincing oneself of something.

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$\operatorname{id}_G$ is not an element of $G/G$. The elements of $G/G$ are sets of the form $gG$ with $g\in G$. But $gG=G$ for all $g\in G$, so there is only one co-set.

Now you just have to convince yourself that all groups with only one element are isomorphic. That's pretty easy to do.

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Maybe it will help you to first look at a different quotient group, and then look at $G/G$.

Suppose that you have a subgroup $H$ of $G$ which is exactly half the size of $G$. The factor group $G/H$ is the set of cosets of H in G. What are these cosets?

If you multiply everything in $H$ by an element that is in $H$, you get back $H$. If you do the same with some fixed $g_0\in G$ such that $g_0\notin H$, the resulting coset $g_0H=\{g_0h : h \in H\}$ has no elements in common with $H$. (If it did have some $g_0h\in H$, by closure $g_0hh^{-1}=g_0\in H$, a contradiction.) Since $g_0H$ is the same size as $H$, which is half the size of $G$, $H$ and $g_0H$ together make up all the elements of $G$. This means that for any $g'\in G$ you pick, either $g'H=H$ or $g'H=g_0H$; in other words, these are the only two cosets.

So we have that $G/H=\{H, g_0H\}$. The operation for a factor group is $(aH)(bH)=(ab)H$, for any two cosets $aH,bH\in G/H$. So, for example, $(g_0H)(g_0H)=(g_0^2)H$. (For the coset "$H$" it's easiest to think of it as having an invisible $\text{id}_G$, so $(g_0H)(H)=(g_0\text{id}_G)H=g_0H$. Thus we have that $H=\text{id}_{G/H}$.) In particular, $G/H$ is a group with $2$ elements, so we know that it is isomorphic to $\mathbb{Z}_2$ because that is the only group of order $2$.

Now let's try doing it with $G/G$.

Well, $gG=G$ for any $g\in G$. So that is the only coset. Thus $G/G=\{G\}$, and $\text{id}_{G/G}=G$. It has only one element, so it must be isomorphic to the only group with one element, the trivial group.

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Thank you. I'll spend some time looking over this. –  dresserse Dec 29 '12 at 22:49
    
The last line is the most clear answer here for sure! –  Squirtle Apr 30 '13 at 3:08

$G/G$ has only one element, which is $G$. Therefore this group is trivial.

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Thanks. I understand $ G/G = \{\text{all elements in G}\} = G $ but how does $ G/G = id_G $? –  dresserse Dec 28 '12 at 21:13
3  
There's a difference between $G$ and {$G$}. Answer says/means {$G$}; you $G$. –  gnometorule Dec 28 '12 at 21:17
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@Peterundergrad $G=\text{id}_{G/G}$, not $\text{id}_G$, and $G/G\not= \{\text{stuff in }G\}$, $G/G=\{gG:g\in G\}=\{G\}$. –  Alexander Gruber Dec 28 '12 at 21:41
    
@AlexanderGruber Thank you. –  dresserse Dec 29 '12 at 22:44

Consider the zero homomorphism $\phi:G\rightarrow G$. Then the $\ker\phi=G$, $\mathrm{Im}\phi =\left\{id_G\right\}$ and from the first theorem of Isomorphism $\frac{G}{\ker\phi}\cong \mathrm{Im}\phi$ which of course implies that $G/G\cong \left\{id_G\right\}$

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Correct, concise answer. Nice. –  andybenji Dec 28 '12 at 21:16
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And probably best explanation for OP who confuses $G$ with {$G$}, and equal with isomorphic. –  gnometorule Dec 28 '12 at 21:23
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Except that he is just starting to learn about quotient groups, does he know what a kernel is, or the relationship between kernel and quotients? If he doesn't understand what $G/G$ is, does he understand the first isomorphism theorem? –  Thomas Andrews Dec 28 '12 at 21:30
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@Thomas Andrews: I was being sarcastic...good, clean answer, but no good given context. –  gnometorule Dec 28 '12 at 21:37
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@gnometorule I got your sarcasm and upvoted. Maybe it's somehow related to the fact that I'm Russian ) –  Dan Shved Dec 28 '12 at 22:09

If G is a group and N is normal in G, then G/N is the quotient group. G/N as a group consists of cosets of the normal subgroup N in G and these cosets themselves satisfy the group properties because of normality of N. Now G is clearly normal in G. Hence G/G consists of the coset that is all of G. Thus this group has only one element, thence it must be isomorphic to the identity group

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$G/G=id_G$ is false. It should be $G/G=\{[id_G]\}$ And actually you could use any element, where the [] mean the equivalence class. Saying G/G=all elements in G is also false, it should be that G/G is a group containing only one element, that is a set that contains all elements in G.

If we go to the definition of quotient group $xy^{-1}\in G\forall x,y\in G$, then, as all elements are related to each other and there's only one equivalence class, therefore the group G/G has one element. As it has one element and the identity must be in the group, then that element is the identity, but it's the identity of the group G/G, not the identity of the group G. The isomorphism G/G->{1} is trivial.

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