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Thread: Normal Subgroup

  1. #1
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    Normal Subgroup

    Prove that A4(+)Z3 has no subgroup of order 18.

    I know A4 does not have a subgroup of order 6 because the converse of Lagrange thereom is false. So my initial notion is that because 6 is a multiple of 18, it cannot exist. Any thoughts on how to make this a proof?

    Thanks All.
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  2. #2
    MHF Contributor Swlabr's Avatar
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    Quote Originally Posted by RoboMyster5 View Post
    Prove that A4(+)Z3 has no subgroup of order 18.

    I know A4 does not have a subgroup of order 6 because the converse of Lagrange thereom is false. So my initial notion is that because 6 is a multiple of 18, it cannot exist. Any thoughts on how to make this a proof?

    Thanks All.
    Yeah, that's the idea. You can use the fact that if $\displaystyle G \cong A_4 \times C_3$ and $\displaystyle H \leq G$ then $\displaystyle H \cong H_1 \times H_2$ where $\displaystyle H_1 \leq A_4$ and $\displaystyle H_2 \leq C_3$. Essentially, that every subgroup of $\displaystyle A_3 \times C_3$ is a direct product of subgroups from these respective groups.

    However, you should probably look at your proof of $\displaystyle A_4$ not having a subgroup of order 6 as $\displaystyle |A_4| = 12$, and 6 divides 12 (of course, I may be getting the wrong end of your stick here)...(if I was proving it, I'd look at what the elements of [tex]A_4[\math] look like).

    Why is your thread called "normal subgroups"?
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  3. #3
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    Whats C3? Thats the chapter the problem is in so I thought maybe somehow normal sbgroups was suppose to be involved.
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  4. #4
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    Quote Originally Posted by Swlabr View Post
    Yeah, that's the idea. You can use the fact that if $\displaystyle G \cong A_4 \times C_3$ and $\displaystyle H \leq G$ then $\displaystyle H \cong H_1 \times H_2$ where $\displaystyle H_1 \leq A_4$ and $\displaystyle H_2 \leq C_3$. Essentially, that every subgroup of $\displaystyle A_3 \times C_3$ is a direct product of subgroups from these respective groups.


    Why is this true in this case? This is definitively not so in other cases, for example: in the Klein group $\displaystyle \mathbb{Z}_2 + \mathbb{Z}_2$, the sbgp. $\displaystyle <(1,1)>=\{(0,0),(1,1)\}$ is not the direct sum of sbgps. of $\displaystyle \mathbb{Z}_2$

    Tonio



    However, you should probably look at your proof of $\displaystyle A_4$ not having a subgroup of order 6 as $\displaystyle |A_4| = 12$, and 6 divides 12 (of course, I may be getting the wrong end of your stick here)...(if I was proving it, I'd look at what the elements of $\displaystyle A_4$ look like).

    Why is your thread called "normal subgroups"?
    .
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  5. #5
    MHF Contributor Swlabr's Avatar
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    Quote Originally Posted by tonio View Post
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    hmm...good point, well made.
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    Quote Originally Posted by RoboMyster5 View Post
    Prove that A4(+)Z3 has no subgroup of order 18.

    I know A4 does not have a subgroup of order 6 because the converse of Lagrange thereom is false.


    Well, that is not a reason but in fact a consequence: $\displaystyle A_4$ has no sbgp. of order 6 because it has none (Add proof here)...

    Here's an idea to prove what you want: suppose $\displaystyle H\,,\,|H|=18$ is a sbgp. of $\displaystyle G=A_4\times C_3$.

    Since $\displaystyle |HA_4|=\frac{|H|\cdot |A_4|}{|H\cap A_4|}\le 36$, we get that $\displaystyle |H\cap A_4|\ge \frac{18\cdot 12}{36}=6\Longrightarrow$ The only option is $\displaystyle |H\cap A_4|=6$ (why??) , which contradicts that $\displaystyle A_4$ has no sbgp. of order 6 .

    Tonio


    So my initial notion is that because 6 is a multiple of 18, it cannot exist. Any thoughts on how to make this a proof?

    Thanks All.
    .
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  7. #7
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    I'm still not comprehending what C3 is, but I am following your idea with the orders.
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  8. #8
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    Quote Originally Posted by RoboMyster5 View Post
    I'm still not comprehending what C3 is, but I am following your idea with the orders.

    $\displaystyle C_3$ is the cyclic group of order 3. It is the same (or better, isomorphic to) as $\displaystyle \mathbb{Z}_3$, but the notation $\displaystyle C_3$ is used when we want to remark the group is a multiplicative one, as opposed to the additive $\displaystyle \mathbb{Z}_3$

    Tonio
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  9. #9
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    Oh okay now this makes a lot more sense, wow thank you so much.
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