- #1
lemonthree
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- 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.
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.