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Group isomorphism proof

lemonthree

Member
Apr 11, 2016
51
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
Jun 20, 2014
1,929
Yes, the order of the statements is logically correct.