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# FP3 Group Isomorphisms (MEI)

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1. This question often comes up, where you are given two groups and have to show the isomporphism.

The identity is fairly obvious but I'm not sure how to find the others.

At the moment I just find an element which generates the whole group if it's cyclic, then find out what I have to do to that element to generate the other elements, and then do the 'same' thing to the element in the other group with can generate the whole group sorry if that's confusing but it seems to work but I don't know why/if there's a logical way to do it? thanks.
2. 4 (i) You need to find an element which generates the whole group e.g. 3 will generate G (check this) since it has order 6. You can use Lagrange's theorem if you wish.

(ii) The proper subgroups must contain the identity which is 1. Make sure the subgroups are also closed and the inverse is in the group e.g. the inverse of 2 is 2x=1 mod 7 --> x=4 so {1,2,4} is a proper subgroup (check it is closed).

(iii) The identity must be mapped to identity. You can use (i) to map 3 to whatever the generator of H is. Since it is a homomorphism it should map the element 3^i of G to the element ...^i of H.
3. (Original post by Math12345)
4 (i) You need to find an element which generates the whole group e.g. 3 will generate G (check this) since it has order 6. You can use Lagrange's theorem if you wish.

(ii) The proper subgroups must contain the identity which is 1. Make sure the subgroups are also closed and the inverse is in the group e.g. the inverse of 2 is 2x=1 mod 7 --> x=4 so {1,2,4} is a proper subgroup (check it is closed).

(iii) The identity must be mapped to identity. You can use (i) to map 3 to whatever the generator of H is. Since it is a homomorphism it should map the element 3^i of G to the element ...^i of H.
Thank you! Sorry I forgot to specific that (iii) was the part I needed help with but your explanations of the other two parts were useful too
4. More questions (didn't seem worth making a new thread)...

Attachment 544671544677
For part (iv), the mark scheme goes from bc=b(ba), to bc=ea. I don't understand where they got b(ba) = ea from?

And for part (vii), I missed the last two subgroups listed in the markscheme, is there a method to find them? Usually I just use each element as a generator but I don't think the last two subgroups can't be obtained like that.

Thanks
Attached Images

5. (Original post by Alex621)
More questions (didn't seem worth making a new thread)...
For part (iv), the mark scheme goes from bc=b(ba), to bc=ea. I don't understand where they got b(ba) = ea from?
b has order 2, so b^2 = e.
6. (Original post by Zacken)
b has order 2, so b^2 = e.
Thanks I didn't read the question carefully enough
7. (Original post by Alex621)
M
And for part (vii), I missed the last two subgroups listed in the markscheme, is there a method to find them? Usually I just use each element as a generator but I don't think the last two subgroups can't be obtained like that.
A single generator method will only find cyclic (sub)groups, and the two you didn't get don't belong in that category.

We know the order of any subgroup must divide the order of T, so it's 1,2,4, or 8.
We can ignore 1 and 8 they are not proper subgroups.

You've identified all the subgroups of order 2, and the first one of order 4 which is generated by F or H

Any other subgroup of order 4, cannot contain F or H. If it did, it would have to contain the subgroup generated by F or H, and will either be the subgroup we've already got, or be the whole of T.

So, any other subgroup comes from the elements {A,B,C,D,E,G}

It's now just a question of being systematic - there are 10 possibilities to check.

E must be in it.

So, lets start by adding A, to get {E,A}
Being systematic we try B, {E,A,B}
AB gives us G, so we must add that {E,A,B,G} and it turns out this is closed, hence a subgroup.

Let's go back a step to {E,A}.
Suppose we add C. to get {E,A,C}
Well AC gives us H. Which means it must contain the whole of the subgroup generated by H, as well as C, giving >4 elements. It must therefore have 8 elements and be the whole of T, so we an ignore the {E,A,C} combination.

Next we try {E,A,D}
Etc.

Bit tedious, and there may be a better way.

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