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In $D_4\times\mathbb Z_2$, find normal subgroups $H$ of orders $2$,$4$, and $8$. For each $H$ describe $G/H$.

I know what makes a subgroup normal, but I don't know how to find one. And what would it mean to describe G/H? I could think of is maybe the left cosets and isomorphic subgroups?

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What's $\,D_4\,$ for you? The dihedral group of order $\,8\,$ or the one of order $\,4\,$? Most probably the former, as the later would make things completely trivial. – DonAntonio Dec 14 '12 at 16:35
it's the one of order 8 – Susan Dec 14 '12 at 16:36
Is it $D_4 \times \mathbb{Z}_2$ or $D_4 \times \mathbb{Z}_4$? You say both. – andybenji Dec 14 '12 at 16:48
Note, also, these are not "normal subgroups of $H$", these are "normal subgroups $H$." They are "normal subgroups of" $G=D_4\times \mathbb Z_4$ (or $G=D_4\times \mathbb Z_2$, depending on which you meant.) – Thomas Andrews Dec 14 '12 at 16:53
It's supposed to be D4 x Z2 – Susan Dec 14 '12 at 16:59

we know that $D_4$ is generated by $a$ and $b$ where $a$ is a permutation of length 4.

  • $H_1= D_4\times \{0\}$ has order 8.

  • $H_2=\{e,a^2b\}\times\{0,2\}$ has order 4.

  • $H_3=\{e\}\times\{0,2\}$has order 2.
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Hints (be sure you can prove the following):

(1) in a direct product, any normal subgroup of one of the factors is normal in the whole group

(2) Besides the above, check the cases where the subgroup is not a subgroup of one of the factors. Sylow subgroups may help here

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