# Group isomorphism proof

#### lemonthree

##### Member
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.

#### Euge

##### MHB Global Moderator
Staff member
Yes, the order of the statements is logically correct.