Is G/G isomorphic to the trivial group? A proof for G/G\cong \{e\}

However, your summary is missing some key information. Here is a more complete summary:In summary, to prove that G/G\cong \{e\}, where \{e\} is the trivial group, we must show that there is a group isomorphism between G/G and \{e\}. This can be done by first establishing that G is the unique left coset of G in G. From this, we can conclude that G/G=\{G\} and since G\lhd G, the quotient group has order |G/G|=1. Then, we define \phi:G/G\to \{e\} as \phi(G)=e, which is a trivially a group isomorphism
  • #1
lemonthree
51
0
Reorder the statements below to give a proof for \(\displaystyle G/G\cong \{e\}\), where \(\displaystyle \{e\}\) is the trivial group.

The 3 sentences are:
For the subgroup G of G, G is the unique left coset of G in G.
Therefore we have \(\displaystyle G/G=\{G\}\) and, since \(\displaystyle G\lhd G\), the quotient group has order |G/G|=1.
Let \(\displaystyle \phi:G/G\to \{e\}\) be defined as \(\displaystyle \phi(G)=e\). This is trivially a group isomorphism and so \(\displaystyle G/G\cong \{e\}\).

I have ordered the statements to what I believe is right but I would just like to check and ensure I'm thinking on the right track.
Firstly, the question wants a proof for a group isomorphism. So we state that the subgroup of G of G is the unique left coset.
Then G/G has got to be {G} since it's the same group G anyway. We know that the order is 1.
Therefore, we define phi to be the proof, and from there we can conclude the group isomorphism.
 
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  • #2
Yes, the order of the statements is logically correct.
 

FAQ: Is G/G isomorphic to the trivial group? A proof for G/G\cong \{e\}

What is the definition of an isomorphic group?

An isomorphic group is a group that has the same structure as another group, meaning that the elements and the operation between them are preserved.

What is the trivial group?

The trivial group, denoted as {e}, is a group that only contains the identity element and has no other elements. It is the simplest and smallest group.

Why is G/G isomorphic to the trivial group?

Since G/G is the quotient group of G by itself, it contains only the identity element. This means that the structure of G/G is the same as the structure of the trivial group, making them isomorphic.

Can you provide a proof for G/G\cong \{e\}?

Proof: Let G be a group and e be the identity element of G. Then, G/G is the set of all left cosets of G with respect to G. Since G is a group, it contains the identity element e. Therefore, every left coset of G is equal to G. This means that G/G only contains the coset G, which is equivalent to the identity element e. Thus, G/G is isomorphic to the trivial group {e}.

Are there any exceptions to G/G\cong \{e\}?

No, there are no exceptions to this statement. For any group G, G/G will always be isomorphic to the trivial group {e}.

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