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Thread: Group homomorphism problem

  1. #1
    Oct 2011

    Group homomorphism problem


    i have this question,

    From my notes these groups are written as addition modulo 10 and 8, but this means that they both have different sizes, so no homomorphism can exist right?

    This makes me think that maybe its supposed to be multiplication modulo 10 and 8, which gives

    Z10 = {1,3,7,9}

    Z8 = {1,3,5,7}





    Y(9)=7 Y(9.7)=Y(3)=3

    Y(x)=(x-1)(x-3) + x works

    Y(7) = 24 + 7 = 7 (since its in Z8)

    Y(9) = 57 = 1

    But if this is the case, is it even possible to find ALL the homomorphisms?

    Do you guys think that this question was just kind of a trick question, to explain why no homomorphism exists between the two groups (under addition modulo) since their sizes are not the same, (would not be 1-1)?
    Last edited by Angela11; May 5th 2012 at 07:02 AM.
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  2. #2
    Member Sylvia104's Avatar
    Mar 2012
    London, UK

    Re: Group homomorphism problem

    Homomorphisms can exist between groups of different orders. The problem given to you is really about the additive groups $\displaystyle \mathbb Z_{10}=\mathbb Z/10\mathbb Z$ and $\displaystyle \mathbb Z_8=\mathrm Z/8\mathbb Z.$ (NB: Homomorphisms need not be isomorphisms. They need not be bijective.)

    Now $\displaystyle \mathbb Z_{10}$ and $\displaystyle \mathbb Z_8$ (meaning the additive groups) are cyclic; therefore any homomorphism $\displaystyle Y:\mathbb Z_{10}\to\mathbb Z_8$ maps the generators of $\displaystyle \mathbb Z_{10}$ to $\displaystyle \mathbb Z_8.$ Now $\displaystyle 1\in\mathrm Z_{10}$ is obviously a generator of$\displaystyle \mathbb Z_{10}.$ You've found the generators of $\displaystyle \mathbb Z_8$ as $\displaystyle \{1,3,5,7\}.$ What are the possible values of $\displaystyle Y(1)?$
    Last edited by Sylvia104; May 5th 2012 at 01:23 AM.
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  3. #3
    Oct 2011

    Re: Group homomorphism problem

    thanks for replying

    ahh sorry I just realised the things i initial posted are the generators of the additive groups like you said! I posted a bunch of useless stuff so i deleted it lol, just trying again now

    so the homomorphism would be

    Y(1) = 1, Y(3) = 3, Y(7) = 5, Y(9) = 7

    Y(1*3) = Y(4) = Y(1^4) = (Y(1))^4 = 4

    Y(1)Y(3) = 1*3 = 4

    and similarly for the other elements, so is that one homomorphism?

    Is it the only homomorphism since its directly mapping the generators to one another? Any other I could think of would map things like 1 to 3 and then the axioms wouldn't hold
    Last edited by Angela11; May 5th 2012 at 01:51 AM.
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  4. #4
    MHF Contributor

    Mar 2011

    Re: Group homomorphism problem

    if Y(1) = 1, then we must have Y(n) = n for all n, since:

    Y(n) = Y(1+1+...+1) (n times)

    = Y(1)+Y(1)+...+Y(1) (n times) since Y is a homomorphism,

    = 1+1+...+1 (n times)

    = n.

    but this won't work. why? consider what happens to Y(8).

    Y(8) = 8 = 0 (in Z8).

    now let's look at Y(3). since Y(n) = n for all n in Z10, we must have Y(3) = 3.

    but in Z10, 3 = 8+5 (since 13 = 3 (mod 10)).

    and: Y(3) = Y(8+5) = Y(8) + Y(5) = 0 + 5 = 5, if Y is to be a homomorphism.

    what went wrong?

    note that whatever we choose for Y(1), it will be SOME element of Z8. and every element of Z8 has order dividing 8.

    so if Y is a homomorphism, then since Y(1) has order dividing 8:

    0 = Y(1)+Y(1)+Y(1)+Y(1)+Y(1)+Y(1)+Y(1)+Y(1) = Y(1+1+1+1+1+1+1+1) = Y(8).

    hence Y(8+8) = Y(8) + Y(8) = 0 + 0 = 0. but in Z10, 8+8 = 6. so Y has to send 6 to 0.

    similarly, Y(2) = Y(6+6) = Y(6) + Y(6) = 0 + 0 = 0.

    so no matter what, any homomorphism Y has to send 2 to 0.

    now Y(2) = Y(1+1) = Y(1) + Y(1) = 0, which means Y(1) has either order 1 or 2 in Z8.

    the ONLY element of order 1 in Z8, is the identity, 0.

    if Y(1) = 0, then Y(n) = Y(1+1+...1) (n times)

    = Y(1)+Y(1)+...+Y(1) (n times)

    = 0+0+..+0 (n times)

    = 0. that is, if Y sends 1 to 0, it sends ALL of Z10 to 0. this is a perfectly good homomorphism (the trivial one, which sends everything to the identity of the target group).

    let's look at our other choice, Y(1) is of order 2. there is just one element of order 2 in Z8, 4.

    so Y(1) = 4 is our other choice. this is also a homomorhpism:


    note that {0,4} is a subgroup of Z8 = {0,1,2,3,4,5,6,7}.

    here is another way to look at it: the image of Y, Y(Z10) must be a subgroup of Z8. this tells us that the order of the subgroup Y(Z10) divides 8, so |Y(Z10)| = 1,2,4 or 8 (by lagrange).

    but ker(Y), the kernel of Y (the elements that Y sends to 0), is a subgroup of Z10, and thus has order 1,2,5 or 10.

    now |Y(Z10)| = |Z10|/|ker(Y)|, or put another way:

    |ker(Y)|*|Y(Z10)| = |Z10| = 10.

    this shows that |Y(Z10)| has to divide 10, as well, so the order of the image group is a common divisor of 8 and 10.

    the only common divisors of 8 and 10 are 1 and 2. if |Y(Z10)| = 1, Y must map everything to 0, since {0} is the only 1-element subgroup of Z8. since everything maps to 0, in particular, Y(1) = 0.

    on the other hand, if |Y(Z10)| = 2, then ker(Y) must be a 5-element subgroup of Z10, and Y(Z10) is a 2-element subgroup of Z8. the only 5-element subgroup of Z10 is {0,2,4,6,8} = <2>. the only 2-element subgroup of Z8 is {0,4}.

    since Y(1) ≠ 0 (it is not in the kernel), it must be that Y(1) = 4 (the only other choice we get).


    in any case, the upshot is that there are just two homomorphisms possible, defined by:

    Y(1) = 0, or
    Y(1) = 2.

    neither of these are isomorphisms. in fact, no isomorphism is possible...because Z10 and Z8 are different sizes.
    Thanks from Angela11
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  5. #5
    Oct 2011

    Re: Group homomorphism problem

    Thanks alot Deveno, this helped me understand the concept very much!
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