- #1
lemonthree
- 47
- 0
Reorder the statements below to give a proof for $$G/G\cong \{e\}$$, where $$\{e\}$$ 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 $$G/G=\{G\}$$ and, since $$G\lhd G$$, the quotient group has order |G/G|=1.
Let $$\phi:G/G\to \{e\}$$ be defined as $$\phi(G)=e$$. This is trivially a group isomorphism and so $$G/G\cong \{e\}$$.
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.
The 3 sentences are:
For the subgroup G of G, G is the unique left coset of G in G.
Therefore we have $$G/G=\{G\}$$ and, since $$G\lhd G$$, the quotient group has order |G/G|=1.
Let $$\phi:G/G\to \{e\}$$ be defined as $$\phi(G)=e$$. This is trivially a group isomorphism and so $$G/G\cong \{e\}$$.
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.