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  1. G is internal direct product of two cyclic groups.
  2. G is isomorphic to an abelian group.
  3. Every subgroup of G is normal.
  4. G is non abelian.

Now, i say that we can take example as U(8) which is not cyclic and also every non identity elemnt in it has order 2( which is prime). But what abelian group can U(8) be isomorphic to?

Cause if i let that G is isomorpic to an abelian group. Then from the fact that every subgroup of an abelian group is normal, We get that option 3 will get correct.

And i hv to select the incorrect option, which lefts us with option 1 and 4th. And i dont have any clue regarding option 1. As per option 4th if i take group as U(8) option 4th gets incorrect as U(8) is abelian.

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  • $\begingroup$ What group is $U(8)$? $\endgroup$ – Arthur Jan 20 '17 at 15:25
  • $\begingroup$ U(8)={1,3,5,7} with relatively prime elements and group with multiplicative operation. $\endgroup$ – Parul Jan 20 '17 at 15:26
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The specific group you call U(8) is isomorphic to $Z^2$x$Z^2$ (also known as the Viergruppe ['four-group']). For this group only 4 is false.

ON THE OTHER HAND:

1 and 4 must be false in general - Counterexample: G could be (isomorphic to) $Z_p$x$Z_p$x$Z_p$ for some prime p.

2 must be false in general - Counterexample: G could be a group of order $p^3$ for some prime p, with presentation { $x$, $y$, $z$ | $x^p$=$y^p$=$z^p$=1, $xy$=$yx$, $xz$=$zx$, $yz$=$zxy$ }.

Off the top of my head, I cannot say whether 3 is true or false in general.

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Consider Klien 4 group, you will see that option 2 is false.

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