Results 1 to 3 of 3
Like Tree1Thanks
  • 1 Post By Deveno

Math Help - Is Z/2Z a subgroup of Z/3Z?

  1. #1
    Senior Member Educated's Avatar
    Joined
    Aug 2010
    From
    New Zealand
    Posts
    433
    Thanks
    12

    Is Z/2Z a subgroup of Z/3Z?

    Is it?

    \mathbb{Z} / 2 \mathbb{Z} = \{ \bar{0}, \bar{1} \} \subset \mathbb{Z} / 3 \mathbb{Z} = \{ \bar{0}, \bar{1}, \bar{2} \}

    It seems to be closed and there does seem to be inverses in it...?

    Like \bar{1} + \bar{1} = \bar{2} = \bar{0} \in \mathbb{Z} / 2 \mathbb{Z}
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Mar 2011
    From
    Tejas
    Posts
    3,401
    Thanks
    762

    Re: Is Z/2Z a subgroup of Z/3Z?

    No.

    For one thing you have to realize that the $\overline{1}$ in $\Bbb Z/2\Bbb Z$ isn't the "same object" as the $\overline{1}$ in $\Bbb Z/3\Bbb Z$. I prefer the notation $[k]_n$ to indicate the modulus when speaking of distinct quotients of $\Bbb Z$.

    By definition:

    $[1]_2 = \{\dots,-5,-3,-1,1,3,5,7,\dots\} = 1 + 2\Bbb Z$, while:

    $[1]_3 = \{\dots,-5,-2,1,4,7,\dots\} = 1 + 3\Bbb Z$.

    As you can see 4 is in the 2nd coset, but not the first, so these aren't even the same subset of the integers.

    Secondly, the GROUP OPERATION in the two sets is completely different (even though we use the same symbol for both).

    In $\Bbb Z/2\Bbb Z$, we have:

    $\overline{k} + \overline{m} = \overline{k+m} \stackrel{\text{def}}{=} k+m\text{ (mod }2)$

    In $\Bbb Z/3\Bbb Z$, we have:

    $\overline{k} + \overline{m} = \overline{k+m} \stackrel{\text{def}}{=} k+m\text{ (mod }3)$.

    Another way to see that $\overline{1}$ is different in the two groups, is that it has order 2 in the first group, but order 3 in the second group. It would be a good idea to convince yourself that if $g \in G$ has order $k$, and $H$ is a subgroup of $G$ with $g \in H$ as well, that $g$ STILL has order $k$ in $H$.

    Finally, Lagrange's theorem states that ANY subgroup of a finite group has order dividing the order of the entire group. As 2 does not divide 3, no group of order 3 can have a subgroup of order 2.

    (And the subset $\{\overline{0},\overline{1}\}$ is NOT closed in $\Bbb Z/3\Bbb Z$: in $\Bbb Z/3\Bbb Z$ we have:

    $\overline{1} + \overline{1} = \overline{2}$ since 2 is congruent to 2 (mod 3), and NOT congruent to (0 mod 3) - which would mean, by definition, that 2 - 0 = 2 is a multiple of 3).
    Thanks from Educated
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Senior Member Educated's Avatar
    Joined
    Aug 2010
    From
    New Zealand
    Posts
    433
    Thanks
    12

    Re: Is Z/2Z a subgroup of Z/3Z?

    Wow thanks a lot!

    That's many ways to answer my question, and really helps. I've been using modulus for a while now and I only just realised they are quotient groups haha
    I've never heard of the modulo being represented as cosets but that's actually a good way to look at it. It makes so much more sense now...

    Our lecturer only said to think of the (mod) as a clock and the numbers just go back to the start after you reach the modded number, and then moves on to teach quotient groups but never made the connection.

    I'll see if I can prove that if g \in H \leq G then g has the same order in H and G. If not I'll probably make another thread
    Last edited by Educated; June 8th 2014 at 02:36 AM.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. when a subgroup is in the center of other subgroup
    Posted in the Advanced Algebra Forum
    Replies: 2
    Last Post: December 14th 2011, 08:52 AM
  2. Replies: 2
    Last Post: March 2nd 2011, 09:07 PM
  3. characterisitic subgroup implies normal subgroup
    Posted in the Advanced Algebra Forum
    Replies: 2
    Last Post: April 8th 2010, 04:13 PM
  4. Centralizer of a subgroup is a subgroup of the main group
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: March 1st 2010, 08:16 AM
  5. Normal subgroup interset Sylow subgroup
    Posted in the Advanced Algebra Forum
    Replies: 0
    Last Post: May 10th 2008, 01:21 AM

Search Tags


/mathhelpforum @mathhelpforum