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Third Isomorphism Theorem

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Feb 15, 2012
Let G be a group and H, J be normal in G with J containing H. I can prove all of the theorem except showing that the homomorphism f: G/H-> G/J defined by f(gH)=gJ is well defined! This means I need to show that gH=bH for b,g in G implies that gJ=bJ.


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MHB Math Scholar
Feb 15, 2012
suppose xH = yH for two elements x,y of G.

then y-1x is in H. but H ⊆ J! so y-1x is in J.

thus xJ = yJ.

perhaps an example will make this clearer:

we'll use an abelian group G, so we don't have to worry about normality.

let G = Z12 under addition mod 12.

let J = {0,2,4,6,8,10}, and let H = {0,4,8}.

clearly J contains H.

the cosets x+J:

J = {0,2,4,6,8,10}
1+J = {1,3,5,7,9,11}

(these cosets have "other names", for example 4+J = J, and 7+J = 1+J).

the cosets x+H:

H = {0,4,8}
1+J = {1,5,9}
2+J = {2,6,10}
3+J = {3,7,11}

notice anything?

when H partitions G, it "respects the partition induced by J", we just chop the cosets by J into SMALLER cosets by H. so:

J = H U (2+H)
1+J = (1+H) U (3+H)

so any two elements in the same coset of H are in the same coset of J (we get cosets of J by "lumping together cosets of H").

specifically the map x+H --> x+J takes:

2+H-->J (and 2 is in J, this works)
3+H-->1+J (and 3 is in 1+J, so this is fine, as well).

you can also look at it this way:

the cosets xJ chop G up into "J sized pieces"(even if J isn't normal).

since H is a subgroup of J, we can, in turn, chop J into "H sized pieces"

and use this to chop the J-pieces xJ into H-pieces x(yH).

since J is BIGGER, G/J is "chunkier" (bigger pieces), while G/H is "finer" (smaller pieces),

and each bigger chunk of G/J is composed of smaller chunks of G/H.

if both subgroups are normal, then we have a group structure on G/J and G/H

and there is a nice relationship between G/J and G/H, the same relationship enjoyed by J and H (the cosets just "magnify it" by a factor of the indices of the respective subgroups involved).
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Feb 15, 2012
Very informative. Thank you.