# Thread: Factor group

1. ## Factor group

Could someone help me on this problem?
I want to figure out which of the following groups is isomorphic to $\displaystyle (Z_4\times Z_{12}) / <(2,2)>$. The following groups are $\displaystyle Z_8, Z_2 \times Z_2 \times Z_2, Z_4 \times Z_2$

I know $\displaystyle <(2,2)>$ has order 6, so the given group has order 48/6=8. I tried to list the elements of this factor group which has order 8, but I was unable to do so since I don't know how to pick the representatives from $\displaystyle Z_4 \times Z_{12}$ so that they will give 5 distinct elements (I know $\displaystyle <(2,2)>$ is the identity in this factor group). I know $\displaystyle Z_8$ is cyclic, so if I can show that the factor group has no element of order 8, then I can eliminate $\displaystyle Z_8$. I also know every non-identity element of $\displaystyle Z_2 \times Z_2 \times Z_2$ has order 2, so if I can show that the factor group has an element of order 4, then I should be able to finish. Any help is appreciated.

2. Originally Posted by jackie
Could someone help me on this problem?
I want to figure out which of the following groups is isomorphic to $\displaystyle (Z_4\times Z_{12}) / <(2,2)>$. The following groups are $\displaystyle Z_8, Z_2 \times Z_2 \times Z_2, Z_4 \times Z_2$

I know $\displaystyle <(2,2)>$ has order 6, so the given group has order 48/6=8. I tried to list the elements of this factor group which has order 8, but I was unable to do so since I don't know how to pick the representatives from $\displaystyle Z_4 \times Z_{12}$ so that they will give 5 distinct elements (I know $\displaystyle <(2,2)>$ is the identity in this factor group). I know $\displaystyle Z_8$ is cyclic, so if I can show that the factor group has no element of order 8, then I can eliminate $\displaystyle Z_8$. I also know every non-identity element of $\displaystyle Z_2 \times Z_2 \times Z_2$ has order 2, so if I can show that the factor group has an element of order 4, then I should be able to finish. Any help is appreciated.

Writing $\displaystyle \mathbb{Z}_4=\{0,1,2,3\}\!\!\!\pmod 4\,,\,\,\mathbb{Z}_{12}=\{0,1,2,\ldots, 11\}\!\!\!\pmod{12}$ , we get $\displaystyle \mathbb{Z}_4\times \mathbb{Z}_{12}\geq <(2,2)>:=\{(2,2)\,,\,(0,4)\,,\,(2,6)\,,\,(0,8)\,,\ ,(2,10)\,,\,(0,0)\}$ .

Now note that $\displaystyle \forall\,b\in\mathbb{Z}_{12}\,,\,\,4b=0,4,8\!\!\!\ pmod{12}$ , so for any element $\displaystyle (a,b)$ in the direct product we get $\displaystyle 4(a,b)$ is one of $\displaystyle (0,0)\,,\,(0,4)\,,\,(0,8)$ ,and all these

elements belong to $\displaystyle <(2,2)>\Longrightarrow$ the maximal order of an element in the factor group is 4...does this now give you an idea what this factor group is?

Tonio
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "http://mathhelpforum.com/advanced-algebra/141841-factor-group.html", "fetch_time": 1529599670000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2018-26/segments/1529267864191.74/warc/CC-MAIN-20180621153153-20180621173153-00111.warc.gz", "warc_record_offset": 189967512, "warc_record_length": 9829, "token_count": 870, "char_count": 2747, "score": 3.765625, "int_score": 4, "crawl": "CC-MAIN-2018-26", "snapshot_type": "latest", "language": "en", "language_score": 0.920965850353241, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00061-of-00064.parquet"}