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Let $G=S_3 \times \mathbb{Z}_4$ be a group and $H= \langle(1 2 3)\rangle \times \langle2\rangle$ and $K=\langle(1)\rangle \times \langle2\rangle$ be subgroups.

  1. Show in two different ways that H normal to G, k is normal to H and K normal to G

  2. Find the elements of H/K and G/H

  3. Find an isomorphic group to G/H, and prove it with the isomorphism theorem

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  • $\begingroup$ thanks for your help....... if im asking is because i dooooon't understand it !!!!!! and i really need it for tomorrow!!! $\endgroup$
    – mariert
    Dec 5 '13 at 3:24
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    $\begingroup$ Of course, but you are expected to at least show what it is you don't understand, where are you stuck, what you've tried to solve the problem, &c. $\endgroup$
    – Pedro Tamaroff
    Dec 5 '13 at 3:37
  • $\begingroup$ first I dont understand my group G I Know that S3= (1)(12)(13)(23)(123)(321) order 6 and Z4=0123 but s3 xZ4 ?? $\endgroup$
    – mariert
    Dec 5 '13 at 3:43
  • $\begingroup$ In your previous question you were explained a list of things. Did you read that carefully? $\endgroup$
    – Pedro Tamaroff
    Dec 5 '13 at 3:44
  • $\begingroup$ I did!... this is for take home test for tomorros and i have my final exam friday!! im trying to understand it!! $\endgroup$
    – mariert
    Dec 5 '13 at 3:48
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  1. To see that $H \triangleleft G$, let $(\sigma, n) \in G$. Show that $(\sigma, n) H = H (\sigma, n)$. By property of direct products, it suffices to show separately that both $\sigma \langle (123) \rangle = \langle (123) \rangle \sigma$ and $n+2 = 2+n$. A similar process nets you normality of the other two.

  2. Elements of $H/K$ are of the form $((1), 0 \bmod 2)$ and elements of $G/H$ are of the form $(\tau \bmod (123), n \bmod 2)$.

  3. Hint: in my definition of $G/H$, $\tau$ is a transposition.

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  • $\begingroup$ $(\tau \circ \langle(123) \rangle, n \bmod 2)$. $\endgroup$
    – mariert
    Dec 5 '13 at 4:38
  • $\begingroup$ Hmm, since $\langle (123) \rangle$ is cyclic, that notation really should be $(\tau \bmod (123), n \bmod 2)$. I've edited my response. $\endgroup$
    – JoeDub
    Dec 5 '13 at 4:51
  • $\begingroup$ there is another way to contact you??? I REALLY NEED HELPPP $\endgroup$
    – mariert
    Dec 5 '13 at 4:56
  • $\begingroup$ I would prefer to keep contact here on StackExchange so that other users can chime in while I'm away. What exactly is it that you're still stuck on? $\endgroup$
    – JoeDub
    Dec 5 '13 at 19:14

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