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I am a newcomer to group theory. I am looking at $C_4$ which has the elements $\{1,a,a^2,a^3\}$

Its subgroups are -

order 4: $\{1,a,a^2,a^3\}$

order 2: $\{1,a^2\}$

order 1: $\{1\}$

Why isn't $\{1,a,a^2\}$ a subgroup?

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Keep in mind that $a^4 = 1$ in this group. Thus, even though $a^3 \cdot a^3 = a^6$ does not appear to be an element of $C_4$, it is secretly $a^4 \cdot a^2 = 1 \cdot a^2 = a^2$. –  Austin Mohr Sep 22 '12 at 18:09
    
It seems you saying $a^3$ is a member of $C_4$... –  Jim_CS Sep 23 '12 at 0:14
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It is a member of $C^4$ as you've defined it. What I mean is that you can take two elements ($a^3$ and $a^3$) and multiply them together to get something that, at first glance, does not appear to belong to the group ($a^6$). –  Austin Mohr Sep 23 '12 at 0:46
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2 Answers 2

up vote 12 down vote accepted

Try multiplying $a$ and $a^2$. Does it lie in your "subgroup"?

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Cheers, this is the level I am operating at the moment :( –  Jim_CS Sep 22 '12 at 17:33
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@Jim_CS That's okay. It's by asking questions that you will learn. –  M Turgeon Sep 22 '12 at 17:35
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Well, because $\,a\cdot a^2=a^3\notin \{1,a,a^2\}\,$ , so it isn't closed under the group operation!

Also, Lagrange's theorem tells us that any subgroup's order must divide the group's order...

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